All that is needed to translate the integrand to obtain
\begin{equation}
\mathrm{PV}\int_{-\infty}^{\infty}\frac{\sin(x+a)}{x}\,dx = \mathrm{PV}\int_{-\infty}^{\infty}\frac{\sin x \cos a + \cos x \sin a }{x}\,dx.
\end{equation}
We already know that
\begin{equation}
\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx = \pi \qquad\text{and}\qquad \mathrm{PV}\int_{-\infty}^{\infty}\frac{\cos x}{x}\,dx = 0.
\end{equation}
Even if you did not know these results, they are easier to deal with than the original integral. Then the rest follows immediately.
Of course, you can evaluate it manually by contour integration. It will be greatly helpful to recall how we evaluated the
$$ \int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx = \pi. $$
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
Consider $f(z) = e^{-ax^2}$, and let $C$ be the rectangular contour with vertices at $-R,R,R+ib$ and $-R+ib$.
As $R \to \infty$, the two sides parallel to the imaginary axis disappear, then we have, by Cauchy's theorem:
$$ 0 = \oint_C f(z) \, dz = \int_{-\infty}^\infty f(x)\, dx + \int_{\infty}^{-\infty} e^{-a(x+ib)^2} \implies\\ \int_{-\infty}^\infty e^{-ax^2 -2 a i x b +a b^2}\, dx = \int_{-\infty}^\infty f(x)\, dx = \sqrt{\frac{\pi}{a}} \implies \\ \int_{-\infty}^\infty e^{-ax^2 -2 a i x b}\, dx = \sqrt{\frac{\pi}{a}}e^{-ab^2} $$
Taking real parts of both sides, the result follows.