You are right about the gain, however there is a small issue with your circuit.
Due to the low values of R4/R5/C2, you will have an undesirable roll off at the low end.
The effective input impedance is set by R4, so the roll off frequency (-3dB) will be:
\$ 1\over{2 \pi R C} \$ = \$ 1 \over{2\pi \ 1k\Omega \ 100nF} \$ = \$ 1591Hz \$ which is way too high for a decent audio amp.
If you change R4/R5 to 10k and 47k, and increase C2 to 1uF or above, then you will have a much lower range. Also don't forget to check the ADC loading at the output also, the same thing can happen there also (see the PIC datasheet for the ADC specs and it's maximum input impedance, as @Andy aka mentions 10k is almost certainly too high, in fact I'm not quite sure what your intentions are with C1 and R6)
Another thing is that if you want to keep the 2.5V offset, then you need to leave C1 out as it will just remove it
One last things is that if you are using a standard electret mic, then 100 Ohms is probably a bit low for R3 (1k - 10k is probably better, but again remember about input/output impedances)
I guess by "complete the feedback loop" you mean "hold the inverting and noninverting inputs at the same voltage". This is basically the op-amp's only goal in life, and given suitable negative feedback, it will accomplish it. If it can't, then it will drive the output into one supply rail or the other attempting to do so.
So, why can IC1 accomplish this, while the astable multivibrator can not? Let's consider the essential components of each:
simulate this circuit – Schematic created using CircuitLab
Now consider the definition of capacitance:
$$ I(t) = C\frac{\mathrm dV(t)}{\mathrm dt} $$
It might make a little more sense algebraically re-arranged:
$$ \frac{\mathrm dV(t)}{\mathrm dt} = \frac{I(t)}{C} $$
That says, "the rate of change of current with respect to time is equal to current divided by the capacitance". So, if you put 1A through a 1F capacitor, voltage changes at a rate of 1V/s. If you increase the current or decrease the capacitance, voltage will change faster. To get voltage to change instantly, you need infinite current or zero capacitance.
For IC1, it's easy for the op-amp to respond to any change in the input. The voltage across a capacitor wants to remain constant -- it takes time and current to change it. If in some instant the input voltage increases by 1V, the output can increase by 1V, and instantly the inverting input also increases by 1V, and the two inputs have the same voltage. Mission accomplished.
But what about IC3? Say the input increases by 1V instantly. What can the opamp do? It can increase the output voltage, but the voltage across C2 (and thus, at the inverting input) can not change instantly. To change it instantly would require infinite current. But that's impossible, because the current the op-amp can drive through the capacitor is limited by R1.
So instead, the op-amp will do the best it can and saturate the output at the positive supply rail. Eventually, it will manage to charge C2 to match the voltage at the input, and the output voltage will go to 0V.
To make an astable multivibrator, you add positive feedback so as the output starts to settle to 0V the input voltage also changes. Thus, IC3 (with positive feedback added) can never accomplish its goal. It's always trying to catch up, and every time it succeeds, it starts another cycle.
Best Answer
The Barkhausen criteria are usually applied to analyze sine wave type oscillator circuits (Wien bridge, etc.) where a small signal, such as thermal noise, is exponentially amplified around the positive feedback loop to create the output signal. Oscillation is inherently a large signal phenomena and in general can't be analyzed using LTI analysis methods, but the Barkhausen criteria let you predict oscillation from the small signal gain and phase behavior.
It's less clear to me how to directly apply such techniques to this relaxation oscillator circuit, as circuits like this don't have any small signal behavior - there are only 2 stable states. It should be fairly obvious, however, that whatever component values you choose the feedback around the loop will eventually be unity and in phase, i.e. when V+ = V-.