From each even term in the exponential series, you get one contribution where the positive and negative powers of $z$ cancel out. This is the middle term of the binomial expansion, and to get the coefficient $a_{-1}$ in the Laurent series you need to sum this over all even terms, which leads to
$$a_{-1}=\sum_{k=0}^\infty\frac1{(2k)!}\binom{2k}{k}A^kB^k=\sum_{k=0}^\infty\frac{(AB)^k}{k!^2}\;.$$
The integral as posed does not exist. No Cauchy principal value will help this. I can, however, pose a problem that is very close that will produce a pretty nifty result.
Consider
$$\oint_C dz \frac{e^{1/z}}{z^2-1} $$
where $C$ is the following contour:
i.e., the circle $|z|=1$ with semicircular notches of radius $\epsilon$ cut into the circle at the poles $z=\pm 1$. By Cauchy's theorem, this integral is zero. However, we can use this fact to deduce a nontrivial integral.
The contour integral is also equal to
$$i \int_{-\pi/2}^{-\epsilon} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{1/(1+\epsilon e^{i \phi})}}{(1+\epsilon e^{i \phi})^2-1} \\ + i \int_{\epsilon}^{\pi-\epsilon} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{1/(-1+\epsilon e^{i \phi})}}{(-1+\epsilon e^{i \phi})^2-1} \\ + i \int_{\pi+\epsilon}^{3 \pi/2} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} $$
Note that the point at which I started and ended the integration on the circle is arbitrary; I chose it so that I was able to encompass the small semicircles in their entirety rather than split them up. Now take the limit as $\epsilon \to 0$, and I can simply express the Cauchy principal value over the circle over any interval of length $2 \pi$. By Cauchy's theorem, we have
$$i PV \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{e^{e^{-i \theta}}}{e^{i 2 \theta}-1} - i \frac{\pi}{2} e + i \frac{\pi}{2} e^{-1} = 0 $$
Now separate out real and imaginary parts. With a little manipulation, we may conclude the following:
$$\int_0^{2 \pi} d\theta \frac{e^{\cos{\theta}} \sin{(\sin{\theta})}}{\sin{\theta}} = \pi \left ( e-\frac1{e}\right ) $$
$$PV \int_0^{2 \pi} d\theta \frac{e^{\cos{\theta}} \cos{(\sin{\theta})}}{\sin{\theta}} = 0$$
Sometimes it is interesting where these contour integrals can take us!
ADDENDUM
It is not trivial that the contour integral above is zero. To show this, we need to expand the integrand in a Laurent series about $z=0$:
$$\frac{e^{1/z}}{z^2-1} = \left ( 1+\frac1{z} + \frac12 \frac1{z^2} + \cdots \right ) \frac1{z^2} \left (1+\frac1{z^2} + \cdots \right ) = \frac1{z^2} + \cdots$$
As there are no terms in $1/z$, the residue of this function at $z=0$ is zero, as is the integral.
Best Answer
Things work very nicely if we take a slightly different approach at the start. (The value of the integral is $2\pi$.) Notice that $$\cos\left(\sin\theta\right)=\text{Re}\left(e^{i\sin\theta}\right).$$ Hence, since $e^{\cos\theta}$ is already real our integral becomes $$\int_{0}^{2\pi}\text{Re}\left(e^{\cos\theta}e^{i\sin\theta}\right)d\theta=\text{Re}\int_{0}^{2\pi}e^{\left(e^{i\theta}\right)}d\theta.$$
Substituting $z=e^{i\theta}$ we have $$\text{Re}\int_{\mathbb{S}^{1}}\frac{e^{z}}{iz}dz=2\pi i \text{Res}\left(\frac{e^{z}}{iz}\right) = 2\pi$$ since the residue is $\frac{1}{i}$.
Hope that helps,