How to calculate an azimuth of a sound source in stereo signal
So here's a diagram:



The signal source is at $(x, y)$, left microphone is at $(-1, 0)$, right microphone is at $(+1, 0)$.  $d=2$ is the distance between the microphones.  $b$ is the distance from the source to the right microphone, and $a$ is the extra distance to the left microphone, which will create additional delay and further decrease the amplitude. The grey circle represents the wavefront as it hits the right microphone.


The delay (time difference) between channels will be $T = \frac a c$, where $c$ is the speed of sound.
Since microphones are pressure sensors, and pressure drops with inverse distance ($1/r$ law), the signal amplitude at the left microphone will be $G = \frac{b} {a+b}$ times the signal at the right microphone.  (It's a level ratio, not a level difference.)


If you know $T$ and $G$ then you can find $b$:
$$b=\frac{G T c}{1-G}$$

Then you know $b$ and $a+b$ and using law of cosines and Apollonius' theorem for the median of the triangle you can find $r$ and $\theta$:

$$r = \frac 1 2 \sqrt{4b^2+4ab+2a^2-d^2}$$
$$b^2=-d r \cos(\theta) + r^2 + \frac{d^2}{4}$$
$$\theta=\arccos\left( \frac{r}{d}+\frac{d}{4r}-\frac{b^2}{dr}\right)$$

Then plug in $a$ and $b$ and simplify:

$$r = \frac 1 2 \sqrt{\frac{2(Tc)^2(G^2+1)}{(G-1)^2} -d^2}$$
$$\theta=\arccos\left( \frac{r}{d}+\frac{d}{4r}-\frac{(G T c)^2}{(1-G)^2 d r}\right)$$

If you only know $T$, you can place the source along a hyperbola (shown in green) with focal points at the microphones.  If you only know $G$, then you can place the source along a circle (shown in orange).  The intersection of those curves is the location of the source:



I would think that accuracy becomes smeared out when the green and orange curves meet at sharp angles.

Of course if you only care about separating sounds by location, you can ignore all this and just work in (delay, ratio)-space.