If $\Delta_{1}$ and $\Delta_{2}$ are two geodesic triangles in $\mathbb{H}_{2}$ with the same interior angles $\alpha$ , $\beta$ and $\gamma$, then there is an isometry of $\mathbb{H}_{2}$ that takes $\Delta_{1}$ to $\Delta_{2}$.
My idea is that one can send corresponding points to corresponding points of the triangles by a Möbius Transformation which is also an isometry of $\mathbb{H}_{2}$. But I am unable to find such a transformation. Also there might be some ambiguity regarding this approach.
Any kind of help would be appreciated.
You might be interested in hyperbolic trigonometry, which has a formula implying that equality of corresponding angles of $\Delta_1$ and $\Delta_2$ implies equality of corresponding side lengths: if the side lengths opposite the angles $\alpha,\beta,\gamma$ are $A,B,C$ then $$\cosh \gamma =\cosh \alpha\cosh \beta-\sinh \alpha\sinh \beta\cos C, $$ and so by solving for $\cos(C)$ we see that $C$ is determined by $\alpha,\beta,\gamma$. Similarly $A$ and $B$ are determined by $\alpha,\beta,\gamma$.
Once you know that, construction of the required isometry is pretty straightforward.
First, the side lengths between the $\alpha$ and $\beta$ vertices of $\Delta_1$ and $\Delta_2$ are both equal to $C$ and so there is an isometry $T : \mathbb H^2 \to \mathbb H^2$ taking $\alpha$ vertex of $\Delta_1$ to the $\alpha$ vertex of $\Delta_2$ and the $\beta$ vertex of $\Delta_1$ to the $\beta$ vertex of $\Delta_2$. This isometry is not unique, in fact there are exactly two isometries which do this: if $L_i$ is the line through the $\alpha$ and $\beta$ vertices of $\Delta_i$ then the two isometries each take $L_1$ to $L_2$, and they differ from each other by reflection across $L_2$. However, only one of these two choices is correct, namely the one which takes the half-plane of $L_1$ that contains the $\gamma$ vertex of $\Delta_1$ to the half plane of $L_2$ that contains the $\gamma$ vertex of $\Delta_2$.
The rest of the work is to check that $T$ does indeed take $\Delta_1$ to $\Delta_2$, and this follows by working your way around the triangles using that corresponding angles and side lengths are equal.
Finally, I'll comment that while every Möbius transformation (of the upper half plane model of $\mathbb H^2$) is an isometry, not every isometry is a Möbius transformation. In fact, the Möbius transformations are exactly the orientation preserving isometries.