The second integral on the RHS is
$$i \rho \int_0^{-\pi} d\theta \, e^{i \theta} \frac{e^{-i t \rho e^{i \theta}}}{\rho e^{i \theta}+i\epsilon} = i \rho \int_0^{-\pi} d\theta \, e^{i \theta} \frac{e^{-i t \rho \cos{\theta}} e^{t \rho \sin{\theta}}}{\rho e^{i \theta}+i \epsilon}$$
The magnitude of the above integral is bounded by
$$\int_0^{-\pi} d\theta \, e^{t \rho \sin{\theta}} = \int_0^{\pi} d\theta \, e^{-t \rho \sin{\theta}}$$
Use symmetry and $\sin{\theta} \ge 2 \theta/\pi$, and the integral is further bounded by
$$ 2\int_0^{\pi} d\theta \, e^{-2 t \rho \theta/\pi} \le \frac{\pi}{t \rho}$$
As $\rho \to \infty$, the integral therefore vanishes as $\pi/(t \rho)$ when $t \gt 0$.
When $t \lt 0$, however, the above bounds do not apply; rather, we must close above the real axis. because there are no poles there, the integral is zero. Thus, the step function.
If $\log z$ is interpreted as principal value
$${\rm Log}z:=\log|z|+i{\rm Arg} z\ ,$$
where ${\rm Arg}$ denotes the polar angle in the interval $\ ]{-\pi},\pi[\ $, then the integral in question is well defined, and comes out to $-2\pi i$. (This is the case $\alpha:=-\pi$ in the following computations).
But in reality the logarithm $\log z$ of a $z\in{\mathbb C}^*$ is, as we all know, not a complex number, but only an equivalence class modulo $2\pi i$. Of course it could be that due to miraculous cancellations the integral in question has a unique value nevertheless. For this to be the case we should expect that for any $\alpha\in{\mathbb R}$ and any choice of the branch of the $\log$ along
$$\gamma:\quad t\mapsto z(t):=e^{it}\qquad(\alpha\leq t\leq\alpha+2\pi)$$ we obtain the same value of the integral. This boils down to computing
$$\int_\alpha^{\alpha+2\pi}(it+2k\pi i)\>ie^{it}\>dt=-\int_\alpha^{\alpha+2\pi}t\>e^{it}\>dt=2\pi i\>e^{i\alpha}\ .$$
During the computation several things have cancelled, but the factor $e^{i\alpha}$ remains. This shows that the integral in question cannot be assigned a definite value without making some arbitrary choices.
Best Answer
Following the recommendation of @honey kumar you can expend the integrand into the series. The integrand is analytic in all points of the unit circle, except for $z=0$. So, you can deform the contour into a circle of smaller radius $r\to0$ around $z=0$, and then to choose all terms of the integrand expansion into series containing $\frac{1}{z}$, because only these terms will contribute, according to the residual theorem.
$\frac{e^{\frac{1}{z}}}{z(1-qz)}=\frac{1}{z}\left(1+\frac{1}{z1!}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+qz+q^2z^2+q^3z^3+...\right)$
You need to take only simple poles from this expression. It gives:
$I(q)=\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz=\oint_C\frac{dz}{z}\left(1+\frac{qz}{z1!}+\frac{q^2z^2}{z^22!}+\frac{q^3z^3}{z^33!}+...\right)=\oint_C\frac{dz}{z}e^q=2\pi{i}e^q$
An important point:
thanks to @A rural reader who points out that $I(q)=2\pi{i}e^q$ at $|q|<1$ and that at $|q|>1$ the integral is zero.
Indeed, when we squeeze the contour ($|z|=r\to0$) and if $|q|>1$ the integration path will have to cross the point $z=\frac{1}{q}$ first, and only after we can enjoy the expansion of $\frac{1}{1-qz}=1=qz+q^2z^2+...$. When crossing the point $z=\frac{1}{q}$ we will get an additional residual in this point ($=-e^q$) which will cancel the residual at $z=0$.
We can also show that $I(|q|>1)=0$ directly:
$I(q)=\oint_C \frac{e^{\frac{1}{z}}}{z(1-qz)}dz=-\oint_C \frac{e^{\frac{1}{z}}}{qz^2(1-\frac{1}{qz})}dz=$$=-\oint_C\frac{dz}{qz^2}\left(1+\frac{1}{z1!}+\frac{1}{z^22!}+\frac{1}{z^33!}+...\right)\left(1+\frac{1}{qz}+\frac{1}{q^2z^2}+...\right)=0$
because there is no simple poles at $z=0$.