Subgroup generated by commensuration class of an element of a virtually free group Suppose $G$ is a group. Let’s call $a, b \in G$ commensurate, iff $\exists m, n \in \mathbb{Z} \setminus \{0\}$, such that $a^n = b^m$. One can see, that commensuration is an equivalence relationship:
$$a = a$$
$$(a^n = b^m) \to (b^m = a^n)$$
$$((a^n = b^m) \cap (b^p = c^q)) \to (a^{np} = c^{mq})$$
Thus we can divide the elements of a group into commensuration classes (let’s denote a commensuration class of an element $a$ as $CC(a)$.
My question is:

Suppose $G$ is a finitely generated virtually free group, $a \in G$ is an infinite order element. Is it always true that $\langle CC(a) \rangle$ is virtually cyclic?

Note, that we can not omit virtual freedom of $G$, because of the existence of Baumslag-Solitar groups $BS(n, m) = \langle a, t| ta^nt^{-1} = a^m \rangle$.
$a$ is required to be of infinite order because of the existence of the following two facts:

All finite-order elements of a group form a commensurability class

Proof:
If $a$ and $b$ are finite-order elements, then we can take $n$ and $m$ as their respective orders.
If $a^k = e$ and $a^n = b^m$, $b^{mk} = e$.
Q.E.D.

$(C_2 \ast C_3)’$ is non-cyclic free

Proof: follows from Ping-Pong lemma.
 A: Actually, more is true. Suppose that $G$ is a word-hyperbolic group (and your groups are since they are commensurable to finitely generated free groups). Each infinite order element $g\in G$ has 2-point fixed-point set $\Phi_g$ in the Gromov-boundary $\partial G$ of $G$. If $g, h\in G$ are commensurate then $\Phi_g=\Phi_h$. Let $L_g\subset \Gamma_G$ be the union of all complete geodesics in the Cayley graph of $G$ which are forward/backward asymptotic to the set $\Phi_g$. Then all geodesics $L_g$ is within Hausdorff distance $2\delta$ from each other. In particular, $L_g$ is quasi-isometric to ${\mathbb  R}$. Hence, the stabilizer of $\Phi_g$ in $G$ is finitely generated and quasi-isometric to ${\mathbb  R}$, hence, is virtually cyclic.  
All this can be found, for instance, in the book by Bridson and Haefliger "Metric Spaces of Non-Positive Curvature."
The subgroup $\langle CC(g)\rangle$ is generated by elements stabilizing $\Phi_g$, hence, is contained in the stabilizer of $L_g$. (Actually, it is equal to that stabilizer, but you do not need this.)  
To conclude, if $G$ is word-hyperbolic then for every infinite order element $g\in G$, the group $\langle CC(g)\rangle$ is virtually cyclic. 
