EDIT: I had to modify my answer because partial fraction decomposition of the integrand resulted in the difference of two divergent integrals.
Assume that $a \ne b$, and let
$$I(r) = \int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(x-a) \right)}{\pi(x-a)} \frac{\sin\left(2 \pi(x-b) \right)}{\pi(x-b)} \, \mathrm dx, $$ where $r$ is a nonegative parameter that is not equal to $1$.
Using partial fraction decomposition and the trig identity $$\sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha)\sin(\beta),$$ we get
$ \begin{align} I (r) &= \small \frac{1}{\pi^{2}(a-b)} \left( \int_{-\infty}^{\infty}\frac{\sin \left(2 \pi r(x-a) \right)\sin\left(2 \pi(x-b) \right)}{x-a} \, \mathrm dx - \int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(x-a) \right)\sin\left(2 \pi(x-b) \right)}{x-b} \, \mathrm dx\right) \\ &= \small \frac{1}{\pi^{2}(a-b)} \left(\int_{-\infty}^{\infty}\frac{\sin \left(2 \pi ry \right)\sin\left(2 \pi(y+a-b) \right)}{y} \, \mathrm dy - \int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(z+b-a) \right)\sin\left(2 \pi z \right)}{z} \, \mathrm dz\right) \\ &= \small\frac{1}{\pi^{2}(a-b)} \int_{-\infty}^{\infty} \frac{\sin(2 \pi ry) \sin(2 \pi y) \cos\left(2 \pi(a-b\right))+ \sin(2 \pi ry)\cos(2 \pi y)\sin\left(2 \pi(a-b) \right)}{y} \, \mathrm dy \\ & \small- \frac{1}{\pi^{2}(a-b)}\int_{-\infty}^{\infty} \frac{\sin(2 \pi rz) \sin(2 \pi z) \cos\left(2 \pi r(a-b\right))- \cos(2 \pi rz)\sin(2 \pi z)\sin\left(2 \pi r(a-b) \right)}{z} \, \mathrm dz\\ &= \small \frac{ \sin \left(2 \pi(a-b) \right)}{\pi^{2}(a-b)} \int_{-\infty}^{\infty} \frac{\sin(2 \pi ry) \cos(2 \pi y)}{y} \, \mathrm dy + \frac{ \sin \left(2 \pi r (a-b) \right)}{\pi^{2}(a-b)}\int_{-\infty}^{\infty} \frac{\sin(2 \pi z) \cos(2 \pi rz)}{z} \, \mathrm dz. \end{align}$
If $\alpha > \beta \ge 0$, we have $$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin( \alpha x) \cos(\beta x)}{x} \, \mathrm dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin\left((\alpha-\beta)x \right) + \sin\left((\alpha+\beta)x \right)}{x} \, \mathrm dx \\ &= \frac{\pi}{2} \left(\operatorname{sgn}(\alpha-\beta) + \operatorname{sgn}(\alpha+\beta) \right) \\ &= \pi. \end{align} $$
And if $ \beta > \alpha \ge 0$, we have $$\int_{-\infty}^{\infty} \frac{\sin( \alpha x) \cos(\beta x)}{x} \, \mathrm dx =0. $$
Therefore, $$\int_{-\infty}^{\infty} \frac{\sin \left(2 \pi r(x-a) \right)}{\pi(x-a)} \frac{\sin\left(2 \pi(x-b) \right)}{\pi(x-b)} \, \mathrm dx = \begin{cases} \frac{ \sin \left(2 \pi r (a-b) \right)}{\pi (a-b)} &\text{ if } 0 \le r < 1\\ \frac{ \sin \left(2 \pi(a-b) \right)}{\pi (a-b)} &\text{ if } r >1 \end{cases}$$
The case $r=1$ should follow by continuity.