Here's the problem:
An opamp generally has two input pins and an output pin. Lets call them in1, in2 and out. The purpose of the opamp is to set Vout = N * (Vin1 -Vin2), where N is the gain.
That's all very well, but what do we mean by "Vout"? Ideally it should be the voltage between the output pin and ground. The problem is that the opamp doesn't know where ground is; the only references it has are the supply rails, and they could be at almost any voltage relative to ground.
Basically, we've set the opamp an impossible task. The best it can do is to set the output voltage relative to one of the supply rails. That's what most opamps do, so the open loop PSRR is zero for that rail, and the closed loop PSRR is limited to the loop gain.
Hence kwkam's comment. If you want 60dB PSRR at 1MHz, you need 60dB loop gain at1MHz, so you need a bandwidth of 1GHz.
Now the good news - if you give the opamp a ground reference, then it's at least theoretically possible for the opamp to set it's output voltage relative to ground.
Let's look at a practical example:
I'm guessing you're working with CMOS, but I'm totally unfamiliar with that, so I hope a BJT example is OK.
The first pic below shows a dead-simple opamp. Cdom sets the dominant pole. It's a bit early in the morning for mental arithmetic, but with the values shown I think unity gain is at about 30MHz and max slew rate is about 10V/uS.
Notice that the feedback current through Cdom is proportional to the voltage between the output and the negative supply rail, so the output voltage is referenced to the negative rail and PSRR is lousy as usual.
In the second picture another capacitor, Ccomp, has been added for compensation. This feeds current into the mirror proportional to the voltage between ground and the negative rail.
Ignoring the input signal for a moment, we see that the current through Cdom must equal the current through Ccomp, so the voltage across them must be the same i.e. Vout = Vground irrespective of the negative supply voltage. Presto - much better PSRR, limited only by component matching, parasitics and Murphy's law. Even if you only get matching to 10%, that still gives a 20dB improvement in PSRR, and reduces your bandwidth requirement ten-fold.
IIRC, there are some commercial opamps that have a ground pin. It may be worth looking at their schematics to see what they do with it. There's probably other, smarter ways to compensate PSRR than what I've shown.