I think you may be overcomplicating this a little bit - why don't you just connect the electret microphone (observing polarity) to the microphone wires via a 3.3Kohm resistor and see what happens.
Your circuit won't work as it stands because the signal you are trying to put back onto the line gets fed back to the microphone via the 10k and it'll all end up very low gain in my humble opinion.
I don't think there is a substitute for trying it out and if it's not loud enough, putting a 10uF capacitor across the 3k3 resistor I mentioned and if that doesn't sound like the right volume maybe drop the 3k3 to 1kohm.
As for matching the impedance - this won't make a great deal of difference to the kit that recieves the signal. It's likely got an unbalanced input and is probably not differential so I don't think you need to jump thru hoops on this.
If you still believe the mic on its own won't work, try this: -
1) OA2's +ve input is just a high input impedance as OA1's, so the extra amp gains you nothing.
2) 100k is good for typical circuits, many people also use 10k.
3) The effect of C1 would only be noticeable in very high quality applications, or if you had very long leads on R1 and R2. It doesn't hurt to put it in though.
Don't use C2. A direct connection of a capacitor to ground at the output makes most opamps go unstable. Again, the effect of a properly connected C2 would only be noticeable in the highest quality work. Put a resistor, 100ohm to 1k, between the opamp output and the capacitor to ensure stability.
4) You could use a pot, but why? If you want an adjustable offset in service, then fair enough. However you do not have DC coupled output gain, so that doesn't seem to be the case.
It should not short rails when turned all the way if connected correctly. That is, one end of track to gnd, the other end of track to +ve supply, the wiper as output to a high impedance load.
5) C3 together with the impedance it's feeding (in this case R3) should have a time constant >> the period of the lowest frequency signal. So C3 > 1/(f.R3) (purists please note I've rolled the 2pi into the difference between > and >>)
If you were doing this sum with R1R2 providing the offset directly without OA1 buffering, then instead of R3, you would use the parallel combination of R1 and R2.
6) Your input signal should be from a lower impedance than R3. In this case, the sound card output is nominally zero, and in practice a few ohms. A rule of thumb in audio like this is that 10k input impedance is plenty high enough to be driven by any audio source.
7) As before, it's audio, and opamps, anything in the 10k to 100k range will work OK.
8) Yes you can, but we generally don't do it like that, for too many reasons that I won't go into now.
The standard way to achieve variable output is to use a fixed gain stage, preceded by a potentiometer. This does have higher noise at low output than the way you suggest, but that's acceptable in most applications. Because you are posting here, I assume you aren't attempting world-beating performance.
9) The C4.R5 time constant gives you the frequency for which the gain is 3dB higher than the DC gain, often called the corner, or break frequency. f3dB = 1/(2.pi.R.C)
10) I'd be cautious about the 5v USB supply. Do filter it with an RC (small R, big C), or at least suspect that rail first if there's noise.
TL;DR - Drop OA1. Replace R3 with R1 and R2. Use 'signal in' to feed the top of a 10k log pot, its bottom to ground, and its wiper to C3.
Best Answer
Your circuit has several problems:
simulate this circuit – Schematic created using CircuitLab
Figure 1. A possible solution.
How it works: