Define
$$ f \colon (0,\infty) \to (0,\infty) \, , \, f(a) = \int \limits_\Omega \frac{\mathrm{d}^2 k}{\sqrt{4 a^2 + (\cos(k_x) + \cos(k_y))^2}} \, .$$
Then (assuming $\Delta, t > 0$) your integral is
$$ \frac{1}{8 \pi^2 t} f \left(\frac{\Delta}{4 t}\right) . $$
Due to the symmetry of the integrand we can integrate over $\{(k_x,k_y) \in \Omega :k_x > k_y\}$ twice and use a trigonometric addition formula to find
\begin{align}
f(a) &= \int \limits_0^\pi \mathrm{d} k_y \int \limits_0^{\pi - k_y} \mathrm{d} k_x \, \frac{1}{\sqrt{4 a^2 + (\cos(k_x) + \cos(k_y))^2}} \\
&= \int \limits_0^{\pi/2} \mathrm{d} k_y \int \limits_{k_y}^{\pi-k_y} \mathrm{d} k_x \, \frac{1}{\sqrt{a^2 + \cos\left(\frac{k_x + k_y}{2}\right) \cos\left(\frac{k_x - k_y}{2}\right)}} \, .
\end{align}
The change of variables $k_x = t+s \, , \, k_y = t-s$ yields
\begin{align}
f(a) &= 2 \int \limits_0^{\pi/2} \mathrm{d} t \int \limits_0^t \mathrm{d} s \, \frac{1}{\sqrt{a^2 + \cos^2(t) \cos^2(s)}} = \int \limits_0^{\pi/2} \mathrm{d} t \int \limits_0^{\pi/2} \mathrm{d} s \, \frac{1}{\sqrt{a^2 + \cos^2(t) \cos^2(s)}} \\
&= \int \limits_0^{\pi/2} \mathrm{d} t \, \frac{1}{\sqrt{a^2 + \cos^2 (t)}} \int \limits_0^{\pi/2} \mathrm{d} s \, \frac{1}{\sqrt{1 - \frac{\cos^2(t)}{a^2 + \cos^2(t)} \sin^2(s)}} = \int \limits_0^{\pi/2} \frac{\operatorname{K}\left(\frac{\cos(t)}{\sqrt{a^2 + \cos^2(t)}}\right)}{\sqrt{a^2 + \cos^2(t)}} \, \mathrm{d} t \\
&= \int \limits_0^{1/\sqrt{1+a^2}} \frac{\operatorname{K}(u)}{\sqrt{1-u^2}\sqrt{1-(1+a^2)u^2}} \, \mathrm{d} u \equiv g\left(\frac{1}{\sqrt{1+a^2}}\right) \, ,
\end{align}
where $\operatorname{K}$ is the complete elliptic integral of the first kind, the last step follows from the substitution $u = \frac{\cos(t)}{\sqrt{a^2+\cos^2(t)}}$ and $g \colon (0,1) \to (0,\infty)$ is defined by
$$ g(r) = \int \limits_0^r \frac{\operatorname{K} (u)}{\sqrt{1-u^2} \sqrt{1-u^2/r^2}} \, \mathrm{d} u = \int \limits_0^1 \frac{r \operatorname{K} (r v)}{\sqrt{1-r^2 v^2}\sqrt{1-v^2}} \, \mathrm{d} v \, .$$
I do not think that there is a nice closed-form expression for $g$, so (apart from numerical methods) we will have to rely on further approximations from now on.
Taylor series lead to
$$ g(r) = \frac{\pi^2}{4} r \left[1 + \frac{3}{8} r^2 + \mathcal{O} (r^4) \right] $$
as $r \searrow 0$ , so
$$ f(a) = \frac{\pi^2}{4 a} \left[1 - \frac{1}{8a^2} + \mathcal{O}(a^{-4})\right] $$
as $a \to \infty$ can be used for $\Delta \gg t$. Note that this series can of course also be obtained directly from the original integral. As for the opposite limit, this question shows that we have
$$ g(r) = \frac{1}{8} \log^2 \left(\frac{1-r}{32}\right) + \mathcal{o} (1) $$
as $r \nearrow 1$, which implies
$$ f(a) = \frac{1}{2} \log^2 \left(\frac{a}{8}\right) + \mathcal{o} (1) $$
as $a \searrow 0$.
