Converse to “free module on locally constant sheaf is quasi-coherent”

Let $X$ be a scheme. Let $\mathcal{E}$ be a sheaf of sets on $X$. Then one can define $\mathcal{O}_X\langle\mathcal{E}\rangle$, the "free module over $\mathcal{E}$", as the sheafification of the presheaf $$U \longmapsto \mathcal{O}_X(U)\langle\mathcal{E}(U)\rangle.$$ The stalk of $\mathcal{O}_X\langle\mathcal{E}\rangle$ at a point $x \in X$ is the free module $\mathcal{O}_{X,x}\langle\mathcal{E}_x\rangle$. It can also be described as $f_! f^{-1} \mathcal{O}_X$, where $f : \operatorname{Ét}(\mathcal{E}) \to X$ is the projection of the étalé space associated to $\mathcal{E}$.

In the case that $\mathcal{E}$ is a constant sheaf $\underline{M}$ with stalks some set $M$, the sheaf of modules $\mathcal{O}_X\langle\mathcal{E}\rangle$ coincides with $\mathcal{O}_X^{\oplus M}$ and is therefore free and in particular quasi-coherent. In the case that $\mathcal{E}$ is a locally constant sheaf, the sheaf of modules $\mathcal{O}_X\langle\mathcal{E}\rangle$ is locally free and therefore too quasi-coherent.

I'm wondering whether the converse holds: Does quasi-coherence of $\mathcal{O}_X\langle\mathcal{E}\rangle$ imply that $\mathcal{E}$ is locally constant?

This looks like a simple statement, but I have been unable to prove it or come up with a counterexample. I know some indications that it's true:

• If $X$ happens to be local as a topological space (spectrum of a local ring), then the converse holds: Using $(\mathcal{O}_X\langle\mathcal{E}\rangle)(X) \cong \mathcal{O}_X(X)\langle\mathcal{E}(X)\rangle$ (this requires locality) one can show that the canonical morphism $\underline{\mathcal{E}(X)} \to \mathcal{E}$ is an isomorphism.

• As a consequence, if $\mathcal{O}_X\langle\mathcal{E}\rangle$ is quasi-coherent, the pullback of $\mathcal{E}$ to any of the $\operatorname{Spec}(\mathcal{O}_{X,x})$ is constant. Thus $\mathcal{E}$ is "constant on all infinitesimal neighbourhoods".

• If $\mathcal{O}_X\langle\mathcal{E}\rangle$ is not only quasi-coherent, but even of finite presentation, then $\mathcal{E}$ is locally constant (with finite stalks).

• Let $j : V \hookrightarrow X$ be the inclusion of an open subset. Let $\mathcal{E}$ be $j_!(1)$, the extension of the terminal sheaf on $V$ by the empty set, explicitly given by $U \mapsto \{ \heartsuit \,|\, U \subseteq V \}$. This sheaf is locally constant iff $V$ is a clopen subset. Now furthermore assume that $X$ is integral. In this case we know that $\mathcal{O}_X\langle\mathcal{E}\rangle = j_!(\mathcal{O}_V)$ (extension by zero this time) is quasicoherent iff $V$ is a clopen subset. Therefore the converse holds in this case.