The problem becomes simpler when you do a unitary variable substitution:
$$ s = \frac{t_1+t_2}{\sqrt{2}} \qquad \sigma = \frac{t_1-t_2}{\sqrt{2}}\\ x = \frac{\omega_1+\omega_2}{\sqrt{2}} \qquad y = \frac{\omega_1-\omega_2}{\sqrt{2}}\\ $$
The double Fourier integral then separates into two easy ones:
$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\text{sinc}[b(\omega_1-\omega_2)]e^{-i\omega_1t_1}e^{-i\omega_2t_2}d\omega_1d\omega_2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\text{sinc}(\sqrt{2}by)e^{-isx}e^{-ity}dx\,dy\\ = \left( \int_{-\infty}^{\infty} e^{-isx} dx\right) \left( \int_{-\infty}^{\infty} \text{sinc}(\sqrt{2}by)e^{-ity}dy\right) $$
These we can do with Mathematica:
Sqrt[2π]*InverseFourierTransform[1, x, s]
(* 2π*DiracDelta[s] *)
Sqrt[2π]*InverseFourierTransform[Sinc[Sqrt[2]*b*y], y, t] // Simplify
(* π*(Sign[Sqrt[2]*b-t] + Sign[Sqrt[2]*b+t])/(2*Sqrt[2]*b) *)
Hence the result is
$$ 2\pi\delta(s) \cdot \frac{\pi[\text{sign}(\sqrt{2}b-t)+\text{sign}(\sqrt{2}b+t)]}{2\sqrt{2}b}\\ = \frac{\pi^2}{\sqrt{2}b}\delta\left(\frac{t_1+t_2}{\sqrt{2}}\right)\left[\text{sign}\left(\sqrt{2}b-\frac{t_1-t_2}{\sqrt{2}}\right)+\text{sign}\left(\sqrt{2}b+\frac{t_1-t_2}{\sqrt{2}}\right)\right]\\ = \frac{\pi^2}{b}\delta(t_1+t_2)\left[\text{sign}\left(b-\frac{t_1-t_2}{2}\right)+\text{sign}\left(b+\frac{t_1-t_2}{2}\right)\right]\\ = \frac{\pi^2}{b}\delta(t_1+t_2)\left[\text{sign}(b-t_1)+\text{sign}(b+t_1)\right] $$
where the last step uses the $\delta$-function to constrain $t_2=-t_1$ and thus to simplify the $\text{sign}$-functions' arguments with $\frac{t_1-t_2}{2}=t_1$.
If you're not too picky about the boundaries where $t_1=\pm b$, then the second half can be written as twice the UnitBox function $\Pi$:
$$ \ldots = \frac{2\pi^2}{b}\delta(t_1+t_2)\cdot \Pi\left(\frac{t_1}{2b}\right) $$
This confirms your target solution.