$\newcommand\al\alpha\newcommand\be\beta\newcommand\th\theta\newcommand\om\omega$Making substitutions $\sin^2\al=x$, $\sin^2\be=y$, $\sin2\th=s$, and $\sin2\om=t$, we rewrite the equality in question in the following "algebraic" way:
$$\int_0^1\int_0^1\frac{dx\,dy}{\sqrt{xy(1-x)(1-y)(1-xy)}}
=\int_0^1\int_0^1\frac{2ds\,dt}{\sqrt{st(1-s)(1-t)(1+s)(1+t)}}.$$
Perhaps this could help, especially because the integrands on the left and on the right now look "more similar" to each other than in the original setting.

**A further comment:** The main difficulty here appears to be with the integral, say $L$, on the left. Expanding $\dfrac1{\sqrt{1-xy}}$ in powers of $xy$:
$$\dfrac1{\sqrt{1-xy}}=\sum_{m=0}^\infty a_mx^my^m$$
with $a_m:=(-1)^m\binom{-1/2}m$ and using the expression of the beta function in terms of the gamma function, we have

$$L=\sum_{m=0}^\infty a_m\Big(\int_0^1 dx\, x^{m-1/2}(1-x)^{-1/2}\Big)^2
=\frac{\pi^2}4\,N,$$
where
$$N:=\sum_{m=0}^\infty\frac1{4^{3m}}\,\binom{2m}m^3.$$
According to Mathematica,
$$N=\frac\pi{\Gamma(3/4)^4}=1.393203929\ldots\quad\text{and hence}\quad L=\frac{\pi^3}{4\Gamma(3/4)^4};$$
however, I do not know how Mathematica does this.

One may also note that $N$ is the expected number of times a symmetric simple random walk on $\mathbb Z^3$ visits its starting point. That is,
$$N=\sum_{m=0}^\infty P(X_{2m}=Y_{2m}=Z_{2m}=0)
=E\sum_{m=0}^\infty 1(X_{2m}=Y_{2m}=Z_{2m}=0),$$
where $X_n:=R_1+\cdots+R_n$, the $R_i$'s are iid Rademacher random variables (with $P(R_i=\pm1)=1/2$), and $(Y_n)$ and $(Z_n)$ are iid copies of $(X_n)$.
It seems amusing that, whereas the dimension $3$ of $\mathbb Z^3$ is odd, $N$ is the square of a certain nice expression.

Closely related to $N$ is Pólya's random walk constant, equal $1-1/N$, which is the probability that the random walk will ever return to its starting point.

The equality
$$N=\frac\pi{\Gamma(3/4)^4}=1.393203929\ldots$$
seems to be proved by Montroll, formulas (6.11), (6.12), (6.1).

Iosif Pinelis's transformation). It turns out that $S$ is the Kummer surface of $E^2$, i.e. the K3 surface birational to $E^2 / \{\pm 1\}$. One can thus find an explicit birational transformation, though it may take some work. How does this arise in your research? $\endgroup$ – Noam D. Elkies Feb 17 at 2:471more comments