This integral can be evaluated with Stokes' theorem; see the solution using the method in this MO post for all the details, as well as the parent MSE post, which relates it to "the Chern number of a vector bundle over a torus". (I am not an expert on the subject!)
This problem can thus be boiled down to showing how to convert OP's integral,
$$\mathcal I = \int_0^{2\pi} \int_0^{2\pi} \frac{-\sin s \cos t - \sin s + \cos t}{\left(3 - 2 \sin s + 2 \cos t - 2\sin s \cos t\right)^{3/2}} \, ds \, dt$$
to $I(1)$ as defined in the linked MSE post.
Denote OP's integrand by $f(\sin s,\cos t)$.
$$\begin{align*}
\mathcal I &= \int_0^{2\pi} \int_0^{2\pi} f(\sin s, \cos t) \, dt \, ds \\
&= \int_{-\pi}^\pi \int_{-\pi}^\pi f(\sin(\pi-s), \cos(\pi+t)) \, dt \, ds \\
&= \int_{-\pi}^\pi \left\{\int_{-\pi}^{-\tfrac\pi2} + \int_{-\tfrac\pi2}^\pi\right\} f(\sin s,-\cos t) \, ds \, dt \\
&= \int_{-\pi}^\pi \left[\int_\pi^\tfrac{3\pi}2 f(\sin(s-2\pi), -\cos t) \, ds + \int_{-\tfrac\pi2}^\pi f(\sin s,-\cos t) \, ds\right] \, dt \\
&= \int_{-\pi}^\pi \int_{-\tfrac\pi2}^\tfrac{3\pi}2 f(\sin s,-\cos t) \, ds \, dt \\
&= \int_{-\pi}^\pi \int_{-\pi}^\pi f\left(\sin\left(s+\frac\pi2\right),-\cos t\right) \, ds \, dt \\
&= \int_{-\pi}^\pi \int_{-\pi}^\pi f(\cos s,-\cos t) \, ds \, dt \\[1ex]
\implies \mathcal I &= \boxed{-4\pi}
\end{align*}$$