The Laplace transform is designed to deal with causal problems in which some input is provided to a linear system at $t=0$, and the system provides a response for $t \gt 0$. There is no response for $t \lt 0$, as the phenomenon being modeled is causal. Thus, when we invert the Laplace transform, we adjust the real part of the line along which we integrate as an expression of this causality.
But why to the right of the rightmost pole? Consider a LT $F(s)$ which has isolated poles in the complex plane and no other singularities. We can compute the ILT using the Residue Theorem by either closing to the left or to the right. If we close to the left, then the contour integral
$$\oint_C dz F(z) e^{z t} = \int_{\sigma-i R}^{\sigma+i R} ds F(s) e^{s t} + i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} F \left ( R e^{i \theta} \right ) e^{R t \cos{\theta}} e^{i R t \sin{\theta}} $$
(NB there is a pair of segments from $\sigma \pm iR$ to the circle $z=R e^{i \theta}$, but one can easily show that the integral over these segments vanishes as $R \to \infty$.)
We want the second integral to vanish as $R \to \infty$ so that we may express the ILT in terms of the residues of the poles of $F$. Note that, when we close to the left, $\cos{\theta} \lt 0$, so that the second integral may vanish only when $t \gt 0$. ($F$ must also satisfy certain conditions as $R \to \infty$, but $t$ must be greater than zero in any case.)
Note that when $t \lt 0$, we may not close the contour to the left. Rather, we must close to the right in order to use the Residue Theorem. Over that semicircle, $\cos{\theta} \gt 0$, so the second integral vanishes only when $t \lt 0$.
Thus, to capture causality, we move the line over which we define the integral for the ILT such that all poles are to the left of the line; in that way $f(t) =0$ when $t \lt 0$, as one would expect.
Let $\chi$ be the characteristic function on the unit interval $T=[-1/2,1/2]$, and take the Fourier transform to be the form
$$F(f)(\xi)=\int_{\mathbb R}f(x)e^{-2\pi i\xi x}dx$$
then
$$F(\chi)(\xi)=\mathrm{sinc}(\xi).$$
With the modulation to translation property of the Fourier transform you get
$$F(e^{2\pi iax}\cdot\chi)(\xi)=\mathrm{sinc}(\xi-a),\quad F(e^{2\pi ibx}\cdot\chi)(\xi)=\mathrm{sinc}(\xi-b),$$
which means $\{\mathrm{sinc}(n-a)\}_{n\in\mathbb Z}$,$\{\mathrm{sinc}(n-b)\}_{n\in\mathbb Z}$ are Fourier coefficients of $e^{2\pi iax}\cdot\chi$ and $e^{2\pi ibx}\cdot\chi$ in their Fourier expansions respectively.
Now since the exponential basis $\{e^{2\pi inx}\}_{n\in\mathbb Z}$ is an ONB on $L^2(T)$, the sum in OP is an $\ell^2$ inner product with respect to this ONB, and with the Plancherel theorem it leads to
$$\sum_{n\in\mathbb Z}\mathrm{sinc}(n-a)\mathrm{sinc}(n-b)=\int_Te^{2\pi i(a-b)x}dx=F(e^{2\pi i(a-b)x}\cdot\chi)(0)=\mathrm{sinc}(b-a)=\mathrm{sinc}(a-b)$$
Best Answer
Applying the Fourier Transform over the time variable $t$, you have that $$\mathcal{F}(Sa(3\pi(t-n)))=\frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} e^{i\omega n}$$ assuming that $Sa(3\pi(t-n))=\frac{\sin(3\pi(t-n))}{3\pi(t-n)}$ in your notation (there are two different $sinc$ functions in the literature, usually for Mathematics and Signal Processing). This means that we can express your equation with the Inverse Fourier Transform as:
\begin{align}\sum_{n=-\infty}^\infty Sa(3\pi(t-n)))&=\sum_{n=-\infty}^\infty \int_{-\infty}^\infty \frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} e^{i\omega n} e^{i\omega t}d\omega \\&= \int_{-\infty}^\infty \frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} \left[ \sum_{n=-\infty}^\infty e^{i\omega n} \right]e^{i\omega t} d\omega\\&\stackrel{*}{=} \int_{-\infty}^\infty \frac{\pi}{3\pi} \mathbb{I}_{\{\omega \in [-3\pi,3\pi]\}} 2\pi \sum_{m=-\infty}^\infty \delta(\omega - 2\pi m) e^{i\omega t} d\omega \\&= \frac{2\pi}{3}\int_{-\infty}^\infty \left[\delta(\omega+2\pi) + \delta(\omega) + \delta(\omega - 2\pi) \right]e^{i\omega t} d\omega \\&= \frac{2\pi}{3}\left[ 1 + 2\cos(2\pi t) \right] \end{align}
where (*) follows from the inverse Fourier Series of signal of $1$ with period $2\pi$, and the last step from the inverse Fourier Transform.