Gain is absolutely the ONLY important part of PSRR. Essentially what you are saying is how much can an op-amp when feeding back a signal cancel out any ripples introduced from the power supply, not from the input of the circuit..
Lets take a simple example: an ideal (infinite open loop gain) voltage follower (output tied directly to the inverting input, fed from the non inverting input). The circuit has a closed loop gain of 1, but the feedback (since the overall gain is SOOO high) will mean that any power supply ripple will be canceled due to the feedback forcing the non inverting and inverting inputs to be in perfect lockstep..
But take the SAME example, but make the OPEN loop gain of the opamp 1, still with closed loop gain of 1, then suddenly the op amp can't keep up with the changes between the non-inverting input and the output-inverting input. And hence all ripple from the power supply would be visible on the output (essentially the op-amp would turn into a noise source with the noise being the coupled power supply ripple)
I understand HOW stevenvh could say that the gain is not meaningful, because he meant CLOSED loop gain... But the gain of question is OPEN loop gain, and YES, that is EVERYTHING in PSRR.
EDIT: And to answer your question, just to follow up slightly here, the PSRR is related to open loop gain, but the more closed loop gain you introduce, the more power supply ripple you will get on the output (hence the 60dB you reference above)
Here is why: Same example I give above, except this time you have a REAL op amp, (finite open loop gain), and resistors in your feedback path, meaning you have a closed loop gain of some value, say 6dB. Since the resistors behave as a voltage divider, the op-amp has to OVERCOMPENSATE for the power supply ripple being fed back to the non-inverting input. If it can only compensate for 100dB of power supply ripple, you will only get 94dB of rejection. The more closed loop gain you introduce, the less of the power supply ripple you are able to reject.
The whole conversation stems from the separate meanings of open loop and closed loop gain.
2nd EDIT: And the way that you get 60dB, or I get my 94dB is that you have to realize you have to convert dB BACK so for example you need to use
$$
20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{6}{20}}\right) = 94 \mathrm{dB}
$$
$$
20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{40}{20}}\right) = 60 \mathrm{dB}.
$$
And YES the other guy who said it should be 1mV not 1µV on Wikipedia is correct.
Best Answer
Rule of electrical life #1. Noise is everywhere.
Noise comes from two sources. From other places within the system, and also from outside the system, eg. the wife's hair-dryer, your cell phone.
Noise is especially prevalent in mixed mode systems. That is, where there is a mixture of digital (Switching) and analog circuits.
Most power supplies these days use digital switching to produce voltages at a minimum of weight and physical volume. As such they tend to be noisier than traditional linear power supplies.
Noise is normally reduced by the liberal application of bypass and decoupling capacitors between the rails. Large values are usually used close to where the power rail enters the PCB, and faster smaller capacitors distributed across the board.
In some cases, when more isolation is required, part of the power rail is split off from the main rail through a small resistor to another bulk capacitor close to the linear circuit it is powering. The resistor capacitor combination forms a low pass filter that significantly removes high frequency noise. However, this method should be used carefully for digital or high frequency linear circuits.
As for differential op-amps. The trick here is to ensure that both inputs are subject to the same amount of noise. Since the circuit subtracts the two inputs, the noise cancels out. To do this properly, the input signal routing should be common to both and if wires are involved they should be twisted together in a twisted-pair.