Since $z$ and $w$ always occur in the same combination, let's write $y=z-\bar w$. The integrand has simple poles at $\xi_n=(n\pi\mathrm i)/(2\beta)$ for $n\in\mathbb Z\setminus\{0\}$, and since $\sinh (n\pi\mathrm i+\xi)=(-1)^n\xi+O(\xi^3)$, the residues at the poles are
$r_n=(-1)^n\frac{\xi_n}{2\beta}\mathrm e^{\mathrm iy\xi_n}\;.$
Since $\omega,z\in\mathcal S_\beta$ ensures $y\in\mathcal S_{2\beta}$, the integrand goes to $0$ exponentially on both sides of the real axis, and depending on whether $y$ has positive or negative real part, we can close the contour of integration with a half-circle at infinity in the upper or lower half-plane, respectively. Assuming $y$ has positive real part, this yields
$ \begin{eqnarray} \frac1{2\pi}\int_{\mathbb R}\frac{\xi}{\sinh{2\beta \xi}}e^{\mathrm iy\xi}\mathrm d\xi &=& \frac{2\pi\mathrm i}{2\pi}\sum_{n\in\mathbb N}(-1)^n\frac{\xi_n}{2\beta}\mathrm e^{\mathrm iy\xi_n} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\sum_{n\in\mathbb N}(-1)^n\mathrm e^{\mathrm iy\xi_n} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\sum_{n\in\mathbb N}(-1)^n\mathrm e^{-n\pi y/(2\beta)} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\sum_{n\in\mathbb N}\left(-\mathrm e^{-\pi y/(2\beta)}\right)^n \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\frac{-\mathrm e^{-\pi y/(2\beta)}}{1+\mathrm e^{-\pi y/(2\beta)}} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\frac{-1}{1+\mathrm e^{\pi y/(2\beta)}} \\ &=& \frac1{2\beta}\frac{\pi}{2\beta}\frac{\mathrm e^{\pi y/(2\beta)}}{(1+\mathrm e^{\pi y/(2\beta)})^2} \\ &=& \frac\pi{4\beta^2}\frac1{(\mathrm e^{\pi y/(4\beta)}+\mathrm e^{-\pi y/(4\beta)})^2} \\ &=& \frac{\pi}{16\beta^2}\left(\cosh\frac{\pi y}{4\beta}\right)^{-2}\;. \end{eqnarray} $
If $y$ has negative real part, the sign from the clockwise integration cancels the sign in $\xi_n$, and the sign in the exponent ends up in the argument of the even function $\cosh$, so the result is the same.