Transmitters
No, there are more than 850, 880, and 940nm transmitters: the IR spectrum ranges from 700nm to 1mm. A distinct set of values is sold, typically in the 700 to 1400 nm range (IR-A), where 850, 880, and 940 are common values. Here's a selection of the Mouser listing of IR emitters. I've used the TSAL series of emitters from Vishay, although I'm not sure that they're in your current, brightness, or bandwidth specs.
The parameter you're interested in for transmitters is the "Relative Radiant Power vs. Wavelength", and for receivers, you want to know the "Relative Spectral Sensitivity vs. Wavelength". For instance, the bell curve in Fig. 9 of the datasheet for the TSAL6100 shows that it has a relative intensity of 1 at 940nm, and outputs about 0.125 times this intensity at 890 nm. That likely means that it's not bright enough to use with an 850 nm detector, and would be iffy at best with an 880nm detector.
Receivers
On the plus side, receivers are usually more generous, for example the TSOP348 detector [picked at random] has a spectral sensitivity of better than 80% for all wavelengths between 850 and 1050nm. Taos Inc. also makes some nice digital, analog, and frequency output detectors for many wavelengths; I've used them with good success before. This will help you if you need to replace a sensor, especially if it's just used as an on/off digital sensor, for instance in a light curtain application, because 80% is pretty close to 100%.
However, that sort of receiver will only tell you about the quantity of light. If you knew that your LEDs were the same brightness (you don't), then you might be able to infer a frequency (i.e. this one is 75% as bright as the 950nm, therefore it's about 820 or 1070nm). You can also determine that an LED is on with just a digital camera, like the one in your cell phone.
Color Sensors
An infrared camera could tell you the wavelength after compensating for temperature, but would not fit most budgets. (Note: These are awesome for determining all kinds of things - Night vision, temperature gradients, etc.)
What you need for that is an color sensor in the infrared range. A color sensor will have multiple narrow-band and/or filtered detectors, so that you can determine the color of the light. See Figure 1, Photodiode Spectral Responsivity" of the datasheet for the TAOS TCS3200D[pdf] for an example (No, it's not going to be a pretty algorithm...). However, you'll notice that the visible light filters stop at about 750nm, and everything goes back to the same curve. Finding a color sensor that works into the infrared range is left as an exercise to the reader, but this sort of IC is what you're looking for.
An alternative to an IR color sensor (which may not exist) would be to use a broadband sensor with a set of infrared transmitting filters tuned to the region of the spectrum which you need. A quick Google search turned up this page, you'll probably find something better.
Distributors:
As for distributors, I find that Mouser has a better selection and cheaper prices on optoelectronics than Digikey.
I've been trying to do a very low light level project myself the past 2 days with photodiodes and phototransistors. This is for people like myself and the original poster who are pushing light detection without a photomultiplier to the limit (below 0.1 mW/cm^2).
I looked at the first receiver module and its minimum irradiance detection was 0.2 mW/m^2 which is about 10,000 times more (less capable) than what discrete photodiodes and phototransistors can do (maybe they meant cm^2 instead of m^2?). Neither are good for really low light levels according to "Art of Electronics" (1 uA per uW of light page 996), totally not capable of getting near what the human eye can do due to leakage current and noise. He describes using photomultipliers which may be required if your light levels are too low. However, in shining light through my fingers in a well-lit room, I am able to see what my eye can't detect on an oscilliscope (with either PhotoDiode or PhotoTransistor).
Assuming his 1 uA per uW is correct, here is an example: a 5 mm photodiodes and phototransistors have an area of 20 micro m^2. So 1 uW/m^2 (1/1000th of noon sunlight) would generate 20 uA (according to Art of Electr) . [[ 1/1000th of noon sunlight is 1 W/m^2 which is about twice as strong as a 20W incadescent light at 1 meter (6W light output into 12 m^2 surface area of a surrounding sphere). ]]
However, my 880nm phototransistor datasheet indicates 600 uA at 1W/m^2 (0.1 mW/cm^2), wich is 30 times more. This assumes all the light is within the active range of the diode's junction.
Sharp has a much better application note, but it seems to be lacking in explaining which design is best for which situations. Figure 13 is most applicable to what original poster and I need, and figure 10B is very interesting but I don't know what they mean by "improves response". http://physlab.lums.edu.pk/images/1/10/Photodiode_circuit.pdf
When used with an op amp, a phototransistor may not capabale of getting as good of a gain as a photodiode for very low light levels because it to uses a "cheap" method of getting its initial gain (transistor instead of op amp). I suspect a photodiode with a JFET op amp (very low input current) would ultimately provide a higher gain with less noise. In any event, the photodiode or phototransistor with the largest optical receiving area might have the best ability to detect low light levels, but that might also increase noise and leakage by a proportional amount and they are usually the underlying problem. So there is a limit to this type of light detection and ideally efficient phototransistors and photodiodes may ultimately be equally good when used with an op amp, but theoretically I suspect photodiode is a little better. It will be the op amp design that matters and I think figure 13 with a JFET op amp and a well-chosen shunt capacitor across the feedback will be best for phototransistor.
For the dual supply op amp, you can use a "lowish" valued resistor pair (two 1k for 10V Vcc to get 5 mA bias) to split the voltage to create a false ground for the +Vin.
I found R=1M for the feedback resistor much better than R=4.7M. Forrest Mimms in his simple opto book used a 10 M with a parallel 0.002uF and a solar cell instead of a phototransistor or photodiode for "extrememly" low light levels" (maybe a solar cell would be better for your application) It seems all PN junctions seem to operate as a solar cell to some extent, as I've read of using clear-cased small signal diodes to detect light. I'm using a regular 830 nm LED as my "photodiode".
The lens angle of whichever 5mm optical diode you use makes a big difference. +/-10 degrees is roughly 4 times more sensitive than +/-20 degrees....if the light source is coming in from less than +/-10 degrees. If the light source is a big area that is +/-20 degrees in front, then it doesn't matter.
I tested the two circuits below. I could detect 0.3V, 5 ms pulses on the phototransistor's Vo which means 0.3 uA which means 0.05 uW/cm^2 if my reading of the datasheet is correct and if it remained linear (big ifs) all the way down to 0.3uA. Maybe it was 5 uW/cm^2. If 0.05 uW/cm^2 is correct, then the off-the-shelf 830 LED was reading down to 0.5 uW/cm^2. I was shining 10 mW 830 nm light through 1 cm of tissue (my finger). I know that if the light levels I was working with were red, it would have been barely visible. The link below shows using 500 M ohm feedback with a photodiode, indicating much lower light levels. Notice the orientation of their photodiode, which is the same as my LED (backwards from most internet links). I got better results this way.
http://www.optics.arizona.edu/palmer/OPTI400/SuppDocs/pd_char.pdf
Best Answer
Probably a phototransistor. It would be best if you could shield it from outside light and attach it closely to the target.
One thing to consider is if the LED is actually illuminated continuously or merely appears to be illuminated continuously. The phototransistor will likely be fast enough to pick up multiplexing that would not be obvious visually. If so then you'd need some kind of filter.
The general approach would be to use a phototransistor plus a resistor (maybe a parallel cap) and a comparator such as an LM393 with a reference formed by another couple resistors.
Using an LED as a photodiode (or a real photodiode) is possible, but the output current is very low compared to a phototransistor and it would thus be more sensitive to electrical noise and might require a bit more complex circuit (such as a transimpedance amplifier).