Fairly simple question: does there exist an area-preserving map from the hyperbolic plane to the Euclidean plane?
If not, does there exist an area-preserving map from an arbitrarily large subset of the hyperbolic plane, to an arbitrarily large subset of the Euclidean plane?
If so, what does the map look like? It would basically be similar to the "Mollweide projection."
There does in fact exist an area-preserving map, as demonstrated in this video at 11:20: the Lambert azimuthal equal-area projection.
The idea is that you take polar coordinates of the hyperbolic plane and map them to polar coordinates of the euclidean plane via a map $(r, \theta) \mapsto (f(r), \theta)$ where $f$ is chosen such that area is preserved.
Let's derive $f$:
Let $h(r)$ be the area of a hyperbolic ball of radius $r$, and $e(r)$ be the area of a euclidean ball of radius $r$.
Assuming that the hyperbolic plane has curvature $-1$, it is true that $h(r) = 2\pi(\cosh(r)-1)$ and $e(r) = \pi r^2$.
We want to ensure $h(r) = e(f(r))$. Substitution and rearrangement yields $f(r) = \sqrt{2\cosh(r)-2}$.
Another area-preserving map, which is analogous to the sinusoidal projection:
Pick any geodesic $g$ in the hyperbolic plane, and mark a special point $O$ on it.
For any point $P$ on the hyperbolic plane, project $P$ perpendicularly onto $g$ to obtain a point $Q$. Let $x$ denote the signed distance $OQ$, let $y$ denote the signed distance $QP$.
Then the map that takes $P$ to the point $(x \cosh y, y)$ is equal-area.
The similarity is obtained as such: If you replace the hyperbolic plane with a sphere, and let $g$ be the equator in particular, and replace $\cosh$ by $\cos$, you get the sinusoidal projection.
Finally, let's do an analogue of the Lambert cylindrical equal-area projection of the sphere, by taking the previous setup and mapping $P$ to the point $(x, \sinh y)$ instead. Again, this map is equal-area.
The analogue: In the spherical case, the map maps $P$ to $(x, \sin y)$.