The answer is no. As Yves de Cornulier pointed out in a comment, this was proved by [Jack Button][1] when $V\cong \mathbb{Z}$. We follow the approach of his proof for the general case.

A subgroup $C$ of a group
$G$ is weakly malnormal if there exists $g \in G$ such that $|C^g ∩ C| < ∞$, where $C^g=gCg^{-1}$.

Suppose that $G = A\ast_V B$ is simple, where $V$ is virtually cyclic, and $V \neq A, B$. By Corollary 2.2 of [Minasyan-Osin][2], if $V$ is weakly malnormal in $G$, then $G$ is either acylindrically hyperbolic or virtually cyclic. In either case, $G$ would not be simple. Acylindrically hyperbolic groups are [SQ-universal][3], hence have many normal subgroups.  
Thus, we conclude that $V$ is not weakly malnormal in $G$. 





Thus, for every $g\in G$, $|V^g\cap V|=\infty$. But $V$ is virtually cyclic, so $[V:V^g\cap V] < \infty$. Hence we get a homomorphism $\varphi: G \to Comm(V) \cong Comm(\mathbb{Z}) \cong \mathbb{Q}^{\times}$, the [abstract commensurator][4]. Since $G$ is simple, $\varphi$ is the trivial homomorphism. Let $Z < V$ be a finite-index copy of $\mathbb{Z}\cong Z$. Then $|Z^g\cap Z|=\infty$, so the homomorphism $g : Z\cap Z^{g^{-1}} \to Z^g\cap Z$, $ a \mapsto gag^{-1}$ must be trivial (since it is trivial as an representative of $Comm(\mathbb{Z})$), which implies that $Z\cap Z^{g^{-1}} = Z^g \cap Z$. Since $G$ is finitely generated, there exists a finite-index subgroup of $Z$ which is normalized (in fact, centralized) by $G$, a contradiction. 


  [1]: https://arxiv.org/abs/1603.05909
  [2]: https://link.springer.com/article/10.1007%2Fs00208-014-1138-z
  [3]: https://en.wikipedia.org/wiki/SQ-universal_group
  [4]: https://arxiv.org/abs/0902.4542