The large majority of regulators will either be damaged and/or will load the output supply if you connect the outputs directly.
As you say you are using a jumper to power only one regulator at a time you could instead place the jumper in the regulator output circuits. This will mean that all 3 regulators are always powered, but the standby current is usually small compared to load current.
Or you could use an input and output jumper per regulator.
If output jumpers are unacceptable you could place an N-Channel MOSFET in the output of each regulator. Source to load, Drain to regulator output, gate to regulator input. When the regulator is powered the MOSFET is switched on. The MOSFET Vgs_th (turn on voltage) needs to be comfortably LESS than the regulator voltage drop in use. The MOSFET Rdson (on resistance) needs to be low enough to only drop minimal voltage when on at full current. eg a 50 milliohm RdsonFET will drop 50 mV at 1 amp.
Use of Schottky diodes on the outputs as Masterleous suggested WILL work BUT the diodes will typically drop 0.3 to 0.6V and the drop will vary with load. Most Schottky diodes at 1A or so are closer to 0.5V+ than 0.3V so this drop is significant and leads to a variability of outp[ut with load. My series MOSFET concept above does the same thing but the MOSFET acts as a "super diode" with very little voltage drop.
Schottky diode voltage drop:
I frequently see Schottky diode forward voltage drops being quoted as "about 0.3 Volts". This CAN be the case, but it's not usual. To get low forward drop you have to design for it. This usually involves operating the diode close to it's maximum reverse voltage ratings and / or using it at far below its peak rated current. Boty these effects are demonstrated in the table below.
The 1N581x series diodes are 1 amp continuous use Schottkky diodes withy reverse voltage ratings of
1N5817 20 Volts
1N5818 30 Volts
1N5819 40 Volts
At 1A the worst case Vf rating at 25C for each of these is typically 0.45V , 0.55V, 0.60V
Exact voltage quoted varies by manufacturer but few or none quote below 0.45V for the 1N5817. So, if you have an eg 15V circuit, the 1N5817 diode will drop about 0.15V less worst case than the 1N5819. That's about 1% of total voltage and in many cases will not make much difference.
BUT I see these diodes or the equivalent SSxx series used in portable lights and solar powered equipment where they may be used in the output of a boost converter providing eg 3V to drive an LED.
Here 0.15 V is 5% of the voltage used to drive the LED. In a simple DC circuit a 1N5817 driving a 3V LED at 1A would need 3.45V input and 1 1N5819 would need 3.6V (worst case in each instance).
1N5817 efficiency is 3/3.45 =~ 87%.
1N5819 efficiency is 3/3.6 = 83%.
For a given amount on input energy the 1N5817 will operate for (87/83-1)x 100 =~ 5% longer.
Energy losses are 13% and 17% so 1N5819 dissipates (17/13-1)x 100 =~ 30% more energy.
In many cases this degree of difference is not important or even trivial.
In others, an extra 5% runtime can be invaluable. By the time you get to worrying about effects of this order you will also be looking at all other areas in a design that cause losses.
You mention that you want fast switching. None of the devices will be able to do that. Any the feedback circuits internally are designed to be slow so that they don't to much incorporate noise. And in fact some of the feedback is thermal feedback. Put another way, the dominant pole is low, 100's of KHz is typically the the fastest response time. This is true whether you are using a disable pin or connecting/opening the load to the source.
The best way to fix this is that you build an external circuit that current steers between two limbs. On one limb will be your LED chain and on the other a dummy load that draws the same amount of current (and hopefully also has similar load characteristics). An example of such a circuit is a differential pair. You should be able to get 10's of ns switching time then.
The challenge with the two limbs will be that the load characteristic differences may disturb the feedback of the LM317 etc. which then reacts slowly. If you can't match the two limbs then I'd suggest building a current mirror that decouples the current output the LM317 etc. so that it only sees a constant load the mirror transistor sees the variability.
You should be able to simulate all of this in LTSPice or similar before building.
The other factors you mention above not show stoppers, so pick what you want.
Best Answer
Assumption: The requirement is for a "regulated power available" indicator, rather than merely a "mains supply on", since the latter would not be affected by the variable voltage of the power supply as mentioned.
LEDs are essentially current-driven devices, not voltage driven. As long as the supply voltage is at least as much as the rated forward voltage of the LED (plus any headroom for current regulation circuitry), and the current through the LED is regulated to a desired value, the LED will be lit at a constant intensity. Typical indicator LEDs commonly are designed for 20 mA current, but will work excellently at 10 mA.
The easiest way of obtaining a constant illumination from an LED across a wide range of supply voltages, is to use a constant current driver circuit.
This can be made using, for example, an LM317 as a constant current source:
Alternatively, use a constant current 2-terminal device such as the SuperTex CL220 or CL2, simply wired in series with the LED. In other words, it is as simple as using a current limiting resistor with the LED, just with one of these parts instead of the resistor.