Power supply with opamp and BJT transistor

Couple of theories here:

• How sure are your power supplies are coming up symmetrically?
  If one rail comes up before the other, you may have non-zero output voltages from the op-amp for a very short period of time.
• Have you implemented all the PCB-layout practices needed for such high-impedances? At minimum, you're going to need guard-rings on all the ultra-high-impedance nodes.
  The National (Now TI) LMC6082 datasheet has a good discussion of what is required to get board-leakage currents low enough to not be a problem.

This will likely not address the possibility that you have dielectric soakage issues, as discussed in @RocketSurgeon's answer.
A good and easy way to test his answer would be to desolder one of the caps on a bad board, and reverse it. If the offset is flipped in the other direction, it's a dielectric soakage issue (because the persistant charge in the cap will have a single polarity). If the offset voltage does not change, the problem is not the capacitor.

One thing I don't see the dielectric soakage issue explaining is why the charge seems to come back when the circuit is unpowered, and go away when it is powered. Since the element that discharges the capacitor is connected continuously across the cap, (e.g C1||R2, C2||R1), the contribution of any current leaking out of the cap should be a constant, and not affected by the supply voltage.

The only thing that comes to mind for me would be that there is something hygroscopic somewhere, and it injects an offset current. When you power the board, it warms up, and drives the moisture out over time. Turn the board off, and it begins to resorb moisture.

One comment I do have is I don't see why you have both SW1A, and SW1B. You can entirely dispose of SW1B. Just tie both of the R/C pairs together, and to the output of the op-amp. When one of the cap/resistor sets is selected, the other will just slowly discharge. As long as one end is floating (which is accomplished by SW1A), the voltage on the other end is irrelevant.

You have wildly complicated your circuit, assuming your symbols mean what the schematic indicates.

First, an instrumentation amp would look like

^{simulate this circuit – Schematic created using CircuitLab}

You show the upper amplifier, which produces OutInv, as being connected to the same ground as R3 and R9. Under these circumstances, OutInv is forced to ground (plus the input offset of the op amp), so you could profitably get rid of the op amp. Furthermore, a gain of 100 is perfectly reasonable from a single op amp, so you could do the whole thing with 1 op amp, as George Herold suggested. A simple version would be

^{simulate this circuit}

This will have a nominal gain of 101. Since an LM358 has a maximum offset voltage of 7 mV, you could have an offset error of 0.7 volts.

Let's assume, though, that the two input grounds are instead a separate ground, isolated from the output ground by some common-mode voltage. In this case, you have connected the upper op amp incorrectly. R1 should be connected to the - input of both amps. However, if you do this, you'll have gain of about 220, rather than 100. The equation for gain for an instrumentation amplifier (assuming R2 equals R3, R4 equals R5, and R6 = R7) is$$G=(1+\frac{2R_2}{R_1})\frac{R_6}{R_5}$$ Alternatively, depending on the output impedance of V1, you could simply use a single op amp set up as a difference amplifier with a gain of 100. You've shown the V1 as a voltage source, so this seems perfectly reasonable.

If you're worried about a stable output, I assume you're worried about noise from the input. This is best handled with an RC filter between V1 and the input.