The $\sin\theta -\sin\varphi$ switches sign if the substitutions $\theta\leftrightarrow\varphi$ are made (a flip along the main diagonal of the square region of integration), and the absolute value stays inert, so we only need to integrate over the triangle with the vertices $(\theta,\varphi)=(0,0), (2\pi,0), (2\pi,2\pi)$:
$$
I\equiv \int_0^{2\pi}d\theta \int_0^{2\pi}d \varphi e^{ia|\sin\theta-\sin\varphi|}
=
2\int_0^{2\pi}d\theta \int_0^\theta d\varphi e^{ia|\sin\theta-\sin\varphi|}
$$
The kernel of the integral also stays the same if we flip along the secondary diagonal of the square region, which is the substitution $\theta'=2\pi-\phi$, $\varphi'=2\pi-\varphi$,
so we are left with integration over the triangle $(0,0), (2\pi,0), (\pi,\pi)$:
$$
I=
4\int_{triangle} d\theta \int d\varphi e^{ia|\sin\theta-\sin\varphi|}
$$
The value of $\sin\theta-\sin\varphi$ switches sign along the line $\varphi=\pi-\theta$,
so this splits the region in a smaller triangle $(0,0), (\pi,0), (\pi/2,\pi/2)$ where $\varphi\le \theta$ and $\sin\theta-\sin\varphi\ge 0$ and a quadrangle $(\pi/2,\pi/2), (\pi,0), (2\pi,0), (\pi,\pi)$ where $\sin\theta-\sin\varphi \le 0$.
Integration over the smaller triangle is split into 2 triangles at $\theta =\pi/2$
and integration over the quadrangle is split into integration over the tilted square $(\pi/2,\pi/2), (\pi,0), (3\pi/2,\pi/2), (\pi,\pi)$ and the triangle $( \pi,0), (2\pi,0), (3\pi/2,\pi/2)$:
$$
I=
4[\int_0^{\pi/2} d\theta \int_0^\theta d\varphi e^{ia(\sin\theta-\sin\varphi)}
+
\int_{\pi/2}^\pi d\theta \int_0^{\pi-\theta} d\varphi e^{ia(\sin\theta-\sin\varphi)}
$$
$$
+
\int_{\pi/2}^\pi d\theta \int_{\pi-\theta}^\theta d\varphi e^{-ia(\sin\theta-\sin\varphi)}
+
\int_\pi^{3\pi/2} d\theta \int_{\pi+\theta}^{2\pi-\theta} d\varphi e^{-ia(\sin\theta-\sin\varphi)}
$$
$$
+
\int_\pi^{3\pi/2} d\theta \int_0^{\theta-\pi} d\varphi e^{-ia(\sin\theta-\sin\varphi)}
+
\int_{3\pi/2}^{2\pi} d\theta \int_0^{2\pi-\theta} d\varphi e^{-ia(\sin\theta-\sin\varphi)}
]
$$
The integral over the tilted square is routed via the subst. $u=(\theta+\varphi)/2, v=(\theta-\varphi)/2$, $\sin\theta -\sin\varphi=2\cos[(\theta+\varphi)/2]\sin[(\theta-\varphi)/2]$, $\varphi=u-v$, $\theta=u+v$, Jacobi determinant 2:
$$I_1\equiv
\int_{\pi/2}^\pi d\theta \int_{\pi-\theta}^\theta d\varphi e^{-ia(\sin\theta-\sin\varphi)}
+
\int_\pi^{3\pi/2} d\theta \int_{\pi+\theta}^{2\pi-\theta} d\varphi e^{-ia(\sin\theta-\sin\varphi)}
$$
$$
=2\int_{\pi}^{2\pi}du \int_0^\pi dv e^{-2ia \cos u\sin v}
=2\int_0^\pi du \int_0^\pi dv e^{2ia \cos u\sin v}
$$
$$
=2\pi \int_0^\pi dv J_0(2a \sin v)
=4\pi \int_0^{\pi/2} dv J_0(2a \sin v)
=2\pi^2 [J_0(a)]^2.
$$
$$
I_2\equiv
\int_0^{\pi/2} d\theta \int_0^\theta d\varphi e^{ia(\sin\theta-\sin\varphi)}
+
\int_{\pi/2}^\pi d\theta \int_0^{\pi-\theta} d\varphi e^{ia(\sin\theta-\sin\varphi)}
$$
$$
+
\int_\pi^{3\pi/2} d\theta \int_0^{\theta-\pi} d\varphi e^{-ia(\sin\theta-\sin\varphi)}
+
\int_{3\pi/2}^{2\pi} d\theta \int_0^{2\pi-\theta} d\varphi e^{-ia(\sin\theta-\sin\varphi)}
$$
$$
=\int_0^{\pi/2} d\theta \int_0^\theta d\varphi e^{ia(\sin\theta-\sin\varphi)}
-
\int_{\pi/2}^0 d\theta \int_0^{\theta} d\varphi e^{ia(\sin\theta-\sin\varphi)}
$$
$$
+
\int_0^{\pi/2} d\theta \int_0^{\theta} d\varphi e^{-ia(-\sin\theta-\sin\varphi)}
-
\int_{\pi/2}^0 d\theta \int_0^\theta d\varphi e^{-ia(-\sin\theta-\sin\varphi)}
$$
$$
=2\int_0^{\pi/2} d\theta \int_0^\theta d\varphi e^{ia(\sin\theta-\sin\varphi)}
+
2\int_0^{\pi/2} d\theta \int_0^{\theta} d\varphi e^{-ia(-\sin\theta-\sin\varphi)}
$$
$$
=2\int_0^{\pi/2} d\theta e^{ia\sin\theta} \int_0^\theta d\varphi e^{-ia\sin\varphi}
+
2\int_0^{\pi/2} d\theta e^{ia\sin\theta} \int_0^{\theta} d\varphi e^{ia\sin\varphi}
$$
$$
=4\int_0^{\pi/2} d\theta e^{ia\sin\theta} \int_0^\theta d\varphi \cos(a\sin\varphi)
$$
$$
=4\int_0^{\pi/2} d\theta \cos(a\sin\theta) \int_0^\theta d\varphi \cos(a\sin\varphi)
+4i\int_0^{\pi/2} d\theta \sin(a\sin\theta) \int_0^\theta d\varphi \cos(a\sin\varphi)
.
$$
$$
I_3\equiv
\int_0^{\pi/2} d\theta \cos(a\sin\theta) \int_0^\theta d\varphi \cos(a\sin\varphi)
=\frac12\int_0^{\pi/2} d\theta \cos(a\sin\theta) \int_0^{\pi/2} d\varphi \cos(a\sin\varphi)
$$
$$
=\frac12[\int_0^{\pi/2} d\theta \cos(a\sin\theta)]^2
=\frac12[\frac12 \int_0^\pi d\theta \cos(a\sin\theta)]^2
=\frac18 [\pi J_0(a)]^2.
$$
$$
I_4\equiv
\int_0^{\pi/2} d\theta \sin(a\sin\theta) \int_0^\theta d\varphi \cos(a\sin\varphi)
=\int_0^{\pi/2} d\theta \sin(a\sin\theta) \int_0^\theta d\varphi [J_0(a)+2\sum_{k\ge 1}J_{2k}(a)\cos(2k\varphi)]
$$
$$
=\int_0^{\pi/2} d\theta \sin(a\sin\theta) [J_0(a)\theta +\sum_{k\ge 1}\frac{J_{2k}(a)}{k}\sin(2k\theta)]
$$
$$
=2\int_0^{\pi/2} d\theta [\sum_{l\ge 0}J_{2l+1}(a)\sin[(2l+1)\theta]] [J_0(a)\theta +\sum_{k\ge 1}\frac{J_{2k}(a)}{k}\sin(2k\theta)]
$$
$$
=2\int_0^{\pi/2} d\theta [\sum_{l\ge 0}J_{2l+1}(a)\sin[(2l+1)\theta]]
J_0(a)\theta
+2\int_0^{\pi/2} d\theta [\sum_{l\ge 0}J_{2l+1}(a)\sin[(2l+1)\theta]]
\sum_{k\ge 1}\frac{J_{2k}(a)}{k}\sin(2k\theta)
$$
$$
=2J_0(a) \sum_{l\ge 0}J_{2l+1}(a)\frac{(-1)^l}{(1+2l)^2}
+2\sum_{l\ge 0}J_{2l+1}(a)
\sum_{k\ge 1}\frac{J_{2k}(a)}{k} \frac{2k(-1)^{}}{(1+2l+2k)(1+2l-2k)}
$$
$$
=2J_0(a) \sum_{l\ge 0}J_{2l+1}(a)\frac{(-1)^l}{(1+2l)^2}
+4\sum_{l\ge 0}J_{2l+1}(a)
\sum_{k\ge 1}J_{2k}(a) \frac{(-1)^{l+k}}{(1+2l+2k)(1+2l-2k)}
$$
$$
=2J_0(a) \sum_{l\ge 0}J_{2l+1}(a)\frac{(-1)^l}{(1+2l)^2}
+4\sum_{s\ge 1} \sum_{l=0}^{s-1} J_{2l+1}(a) J_{2(s-l)}(a) \frac{(-1)^s}{(1+2s)(1+4l-2s)}
.
$$
$$
I_2=4[I_3+iI_4].
$$
$$
I=4[I_1+I_2].
$$
