Let $G\subset PSL_2(\mathbb R)$ be the group generated by the matrices $$a_n=\begin{pmatrix} 1 & 2\cot\frac{\pi}{n}\\0 & 1\end{pmatrix},\; c_n = \begin{pmatrix} \cos\frac{\pi}{n}&\sin\frac{\pi}{n}\\-\sin\frac{\pi}{n}&\cos\frac{\pi}{n}\end{pmatrix}$$ I've seen several references in literature to the fact that every parabolic element (i.e. element of trace $2$) of $G$ is conjugate to a power of $a_n$, all without proof. Is there an easy way to show this, or a proof of this fact in literature?
If we interpret $G$ as acting on the upper-half plane via fractional transformations, the parabolic elements are precisely the elements which fix exactly one point and they are classified up to power by the point they fix. In particular, powers of $a_n$ fix $\infty$. Thus it suffices to show that if $z$ is the fixed point of some parabolic element $g\in G$, then we have some $h\in G$ such that $h(z)=\infty$.
This is going to be an application of the Poincare Polygon theorem in the hyperbolic plane, which you apply by first constructing a fundamental domain.
From your description of the generators, it looks like the fundamental domain is going to be a four-sided polygon obtained by intersecting a fundamental domain for the infinite cyclic group generated by $a_n$ (the vertical strip between $x=\pm \cot(\pi/n)$) and a fundamental domain for the order $n$ finite cyclic group generated by $c_n$ (the angle between two rays emanating from the rotation point $0+1i$, symmetric across the line $x=0$, and subtending an angle $2\pi/n$). This will be a four-sided polygon with one ideal point at $+\infty$.
To apply the Poincare Polygon Theorem you will have to verify that each finite vertex has angle of the form $2\pi/k$ for some integer $k \ge 1$ (which you already know for the vertex at $0+1i$).
Once that's done, the conclusion of the Poincare Polygon Theorem will tell you that all the parabolic points for the group action are the images of the single ideal vertex $+\infty$ under the group action.