Area method in Lobachevskian geometry
Ok, if your question is "Are there proofs by dissection in hyperbolic geometry?" then the answer is very much yes. Probably the first one is due to Gauss.
Let $T(\alpha, \beta, \gamma)$ be the hyperbolic triangle with angles $\alpha$, $\beta$, and $\gamma$. Gauss computes the area of $T(\alpha, \beta, \gamma)$ using the "windmill" figure. That is, Gauss decomposes $T(0, 0, 0)$ into four triangles - the central one is $T(\alpha, \beta, \gamma)$ and the others are $T(\pi - \alpha, 0, 0)$ and so on.
There are also many, many dissection theorems in higher dimensions. There are important conjectures here - see Conjecture 1.3 of Walter Neumann's article.
However, if you are instead asking about "hinged dissections" then I am a stuck. Here is the best I can do.
Liebmann [Nichteuklidische Geometrie, page 43] fixes a gap in Gauss' proof that the area of $T(\pi - \alpha, 0, 0)$ is $\alpha$. I learned this from Coxeter's "Introduction to Geometry" [Figure 16.4a, page 295]. See also Coxeter's article "Angles and arcs in the hyperbolic plane" [last paragraph on page 19].
Here is the figure:
Of course, this is a bit complicated. Here is a somewhat simplified version. We work in the upper half plane model of $\mathbb{H}^2$. We must show that the area of an ideal triangle is finite. It suffices to show that the following region has finite area:
$C = \{ z = x + iy \in \mathbb{H}^2 \mid x \in [1, 2], y \geq 1 \}$
We'll give a dissection of $C$ and reassemble it into $D$:
$D = \{ z \in \mathbb{H}^2 \mid x \in (0, 2], y \in [1, 2] \}$
Since the closure of $D$ is compact, it has finite area and we will win.
We now define
$C_n = \{ z \in C \mid x \in [1, 2], y \in [2^n, 2^{n+1}] \}$
and
$D_n = \{ z \in D \mid x \in [2^{-n}, 2^{-n+1}], y \in [1, 2] \}$
So, $C = \cup C_n$ and $D = \cup D_n$. Also, $C_n$ is isometric to $D_n$ using a power of the map $z \mapsto z/2$, and we are done.
Remarks:
Coxeter uses two families of geodesics to dissect his figure; I instead use one family of geodesics and one family of horocycles.
This example is also helpful for understanding "spiralling geodesics in incomplete finite area hyperbolic surfaces", but that is a story for a different time.
Well, you may get hyperbolic Ceva's theorem using area in the manner similar to Euclidean's one. Namely, using the formula for the area $E$ of a triangle with sides $a,b,c$ and angles $A,B,C$: $\sin \frac{E}2 \cosh \frac{c}2=\sinh \frac{a}2\sinh\frac{b}2\sin C$ (analogue of Euclidean $E=\frac12ab\sin C$) we get the following interpretation of cevian: a line $\ell$ through a vertex $A$ of triangle $\triangle ABC$ is a locus of points $P$ for which the ratio $$\sin \frac{E(PAB)}2\cdot\cosh\frac{BP}2:\sin \frac{E(PAC)}2\cdot\cosh\frac{PC}2$$ is a constant which I denote $\varkappa(\ell;BA,CA)$. The above areas should be oriented but let's ignore this. When are three lines $\ell_a$ through $A$, $\ell_b$ through $B$, $\ell_c$ through $C$ concurrent?
When $\varkappa(\ell_a;BA,CA)\cdot \varkappa(\ell_b;CB,AB)\cdot \varkappa(\ell_c;AC,BC)=1$.
(Again, I ignore the question of concurrent lines without common point which should be treated differently in hyperbolic geometry.) Now if $\ell_a\cap BC=A_1$, we have $$ \varkappa(\ell_a;BA,CA)= \sin \frac{E(A_1AB)}2\cdot\cosh\frac{BA_1}2:\sin \frac{E(A_1AC)}2\cdot\cosh\frac{CA_1}2=\\ \frac{\sinh\frac{BA_1}2\cosh\frac{BA_1}2}{\cosh \frac c2}: \frac{\sinh\frac{CA_1}2\cosh\frac{CA_1}2}{\cosh \frac b2}= \frac{\sinh BA_1}{\cosh \frac c2}: \frac{\sinh CA_1}{\cosh \frac b2} $$ that yields hyperbolic Ceva.