We need not use contour integration to show that
$$\int_0^\infty e^{-|a|t}t^{s-1}\,dt=|a|^{-s}\Gamma(s) \tag 1$$
Rather, we note that the integration path is on the real line. Hence, enforcing the substitution $t\to t/|a|$ yields
$$\begin{align}
\int_0^\infty e^{-|a|t}t^{s-1}\,dt&=\int_0^\infty e^{-t}\left(\frac{t}{|a|}\right)^{s-1}\,\frac{1}{|a|}\,dt\\\\
&=|a|^{-s}\int_0^\infty e^{-t}t^{s-1}\,dt\\\\
&=|a|^{-s} \Gamma(s)
\end{align}$$
as was to be shown.
It remains to be shown that $\int_0^\infty e^{-at}\,t^{s-1}\,dt=a^{-s}\Gamma(s)$ for $|\arg(a)|<\pi/2$. It is to that end that we proceed.
Contour Integral
Let $f(z)=e^{-|a|z}z^{s-1}$. Note that $f(z)$ has a branch point at $z=0$. We choose the branch cut from $z=0$ to $z=-\infty$ along the negative real axis.
Then, $f(z)$ is analytic on this Riemann sheet and Cauchy's Integral Theorem guarantees that for any rectifiable closed curve, $C$, that does not intersect the chosen branch cut
$$\oint_C f(z)\,dz=0$$
We choose $C$ to be the contour as described in the OP. Therefore we write
$$\begin{align}
0&=\oint_C f(z)\,dz\\\\
&=\int_\epsilon^R e^{-|a|x}x^{s-1}\,dx \tag 2\\\\
&+\int_0^{\arg(a)} e^{-|a|Re^{i\phi}}\,(Re^{i\phi})^{s-1}\,iRe^{i\phi}\,d\phi\tag 3\\\\
&+\int_R^\epsilon e^{-|a|te^{i\arg(a)}}\,(te^{i\arg(a)})^{s-1}\,e^{i\arg(a)}\,dt\tag 4\\\\
&+\int_{\arg(a)}^0 e^{-|a|\epsilon e^{i\phi}}\,(\epsilon e^{i\phi})^{s-1}\,i\epsilon e^{i\phi}\,d\phi\tag 5
\end{align}$$
We assume that $|\arg(a)|< \pi/2$, else the integral in $(4)$ does not converge as $R\to \infty$.
Under this assumption, we see that the integral in $(2)$ becomes
$$\lim_{(\epsilon,R)\to (0,\infty)}\int_\epsilon^R e^{-|a|x}x^{s-1}\,dx=\int_0^\infty e^{-|a|x}x^{s-1}\,dx$$
For the integral in $(3)$ we have
$$\begin{align}
\left|\int_0^{\arg(a)} e^{-|a|Re^{i\phi}}(Re^{i\phi})^{s-1}\,iRe^{i\phi}\,d\phi\right|&\le R^{\text{Re}(s)}\int_0^{|\arg(a)|} e^{-|a|R\cos(\phi)}\,d\phi\\\\
&\le R^{\text{Re}(s)}\int_0^{|\arg(a)|} e^{-|a|R(1-2\phi /\pi)}\,d\phi\\\\
&=\frac{\pi}{2|a|}R^{\text{Re}(s)-1}\left(e^{-|a|R\left(1-\frac 2\pi|\arg(a)|\right)} -e^{-|a|R}\right)\\\\
&\to 0\,\,\text{as}\,\,R\to \infty
\end{align}$$
The integral in $(4)$ becomes
$$\lim_{(\epsilon,R)\to (0,\infty)}\int_R^\epsilon e^{-|a|te^{i\arg(a)}}\,(te^{i\arg(a)})^{s-1}\,e^{i\arg(a)}\,dt=-e^{is\arg(a)}\int_0^\infty e^{-at}\,t^{s-1}\,dt$$
And the integral in $(5)$ vanishes as $\epsilon\to 0$.
Putting everything together reveals
$$\int_0^\infty e^{-at}\,t^{s-1}\,dt=\left(e^{i\arg(a)}\right)^{-s}\int_0^\infty e^{-|a|x}x^{s-1}\,dx=a^{-s}\Gamma(s)$$
which was to be shown!
Best Answer
$e^{-|z|^2}$ is not complex differentiable (other than exactly at $0$), so Cauchy's integral theorem doesn't apply to it.
(And of course, if you use $e^{-z^2}$ instead, that will blow up where your large semicircular path crosses the imaginary axis).