Why the Cauchy Principal Value Gives the Proper Value
Your question boils down to the following
$$
\begin{align}
&\mathrm{Im}\left(\mathrm{PV}\int_{-\infty}^\infty\frac{e^{ix}}x\,\mathrm{d}x\right)\\
&=\mathrm{Im}\left(\mathrm{PV}\int_{-\infty}^\infty\frac{\cos(x)}x\,\mathrm{d}x+i\,\mathrm{PV}\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x\right)\tag{1a}\\
&=\mathrm{Im}\left(\lim_{\epsilon\to0}\int_{|x|\gt\epsilon}\frac{\cos(x)}x\,\mathrm{d}x+i\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x\right)\tag{1b}\\
&=\mathrm{Im}\left(0+i\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x\right)\tag{1c}\\
&=\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: $e^{ix}=\cos(x)+i\sin(x)$ and $\mathrm{PV}$ is linear
$\text{(1b)}$: definition of $\mathrm{PV}$ on the cosine integral
$\phantom{\text{(1b):}}$ $\mathrm{PV}$ of a convergent integral is that integral
$\text{(1c)}$: the integral of an odd function over a domain
$\phantom{\text{(1c):}}$ symmetric about the origin is $0$
$\text{(1d)}$: take the imaginary part
Actually, it doesn't even matter what the Principal Value of the cosine integral is. As long as it exists and is real, it is eliminated by taking the imaginary part.
Avoid Singularities Altogether
There is no singularity whatsoever in
$$
\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x\tag2
$$
The way to compute this integral and never come close to a singularity is to note that
$$
\lim_{R\to\infty}\int_{\gamma_R}\frac{\sin(z)}z\,\mathrm{d}z=0\tag3
$$
where $\gamma_R=[-R,R]\cup[R,R-i]\cup[R-i,-R-i]\cup[-R-i,-R]$. This is because there are no singularities inside this contour for any $R$.
Furthermore, the integral vanishes on $[R,R-i]$ and $[-R-i,-R]$ as $R\to\infty$ since $|\sin(z)|\le\cosh(1)$ and $|z|\ge R$ and the length of each path is $1$. Thus, the integral over both paths is no bigger than $\frac{2\cosh(1)}R$.
Discounting the integrals which vanish, $(3)$ becomes
$$
\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x=\int_{-\infty-i}^{\infty-i}\frac{\sin(z)}z\,\mathrm{d}z\tag4
$$
The path of integration on the right side of $(4)$ passes nowhere near a singularity. Use $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$ to evaluate the right side of $(4)$:
$$
\begin{align}
\int_{-\infty-i}^{\infty-i}\frac{\sin(z)}z\,\mathrm{d}z
&=\frac1{2i}\int_{-\infty-i}^{\infty-i}\frac{e^{iz}}z\,\mathrm{d}z-\frac1{2i}\int_{-\infty-i}^{\infty-i}\frac{e^{-iz}}z\tag{5a}\\
&=\frac1{2i}\lim_{R\to\infty}\int_{\gamma_R^+}\frac{e^{iz}}z\,\mathrm{d}z-\frac1{2i}\lim_{R\to\infty}\int_{\gamma_R^-}\frac{e^{-iz}}z\tag{5b}
\end{align}
$$
where $\gamma_R^+=[-R-i,R-i]\cup Re^{i[0,\pi]}-i$ and $\gamma_R^-=[-R-i,R-i]\cup Re^{i[0,-\pi]}-i$. $\text{(5b)}$ follows because the integrals along the huge arcs go to $0$; $e^{iz}$ vanishes exponentially in the upper half-plane and $e^{-iz}$ vanishes exponentially in the lower half-plane. In fact, the integrals along those arcs are bounded by
$$
\begin{align}
\int_0^\pi\overbrace{e^{-R\sin(\theta)+1}\vphantom{\frac RR}}^{\large e^{\pm iz}}\overbrace{\ \frac{R\,\mathrm{d}\theta}{R-1}\ }^{\mathrm{d}z/z}
&\le\frac{2eR}{R-1}\int_0^{\pi/2}e^{-2R\theta/\pi}\,\mathrm{d}\theta\tag{6a}\\
&\le\frac{2eR}{R-1}\frac\pi{2R}\tag{6b}\\[3pt]
&=\frac{e\pi}{R-1}\tag{6c}
\end{align}
$$
Since $\gamma_R^-$ contains no singularities, the integral on the right-hand side of $\text{(5b)}$ is $0$. Since $\gamma_R^+$ contains the singularity at $0$, whose residue is $1$, we get that the integral on the left-hand side of $\text{(5b)}$ is $2\pi i$.
Putting together $(2)-(5)$, we conclude
$$
\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x=\pi\tag7
$$
