The answer is no. As Yves de Cornulier pointed out in a comment, this was proved by Jack Button 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, 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, 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. 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, say by $g_1, \ldots, g_n$, there exists a finite-index subgroup $Z^{g_1}\cap \cdots \cap Z^{g_n}\cap Z \leq Z$ which is normalized (in fact, centralized) by each $g_i$, and hence by $G$, a contradiction.