Is the sheafication of a “presheaf of $\mathcal{O}_X$-modules” an $\mathcal{O}_X$-module?

Let $$(X,\mathcal{O}_X$$) be a ringed space. A presheaf of $$\mathcal{O}_X$$-modules is a presheaf $$\mathcal{F}$$ of abelian groups on $$X$$ such that $$\mathcal{F}(U)$$ is an $$\mathcal{O}_X(U)$$-module for each open $$U\subseteq X$$, and each restriction map of $$\mathcal{F}$$ is linear with respect to the corresponding restriction map of $$\mathcal{O}_X$$. More precisely, the latter condition says: if $$V\subseteq U\subseteq X$$ are open, then $$\alpha(rm)=\beta(r)\alpha(m)$$ for all $$r\in\mathcal{O}_X(U)$$ and $$m\in\mathcal{F}(U)$$, where $$\alpha:\mathcal{F}(U)\to\mathcal{F}(V)$$ and $$\beta:\mathcal{O}_X(U)\to\mathcal{O}_X(V)$$ are the restriction maps.

Is the sheafication of a presheaf of $$\mathcal{O}_X$$-modules necessarily an $$\mathcal{O}_X$$-module?

An $$\mathcal O_X$$-module structure on an abelian pre-sheaf is the same thing as a morphism of abelian pre-sheaves $$\rho_F\colon \mathcal{O}_X\times F\to F$$, making the associativity-diagram commute. Take associated sheaves and the induced morphism: $$\rho^a_F\colon a(\mathcal{O}_X\times F)\to aF$$ and observe that the natural morphism $$a(\mathcal{O}_X\times F)\to \mathcal{O}_X\times aF$$ is an isomorphism since it is an isomorphism on stalks and both are sheaves. Composing $$\rho^a_F$$ with the inverse, we get a morphism of abelian sheaves $$\mathcal{O}_X\times aF\to aF$$ which makes the associativity-diagram commute since it commutes on stalks. Thus, we find the induces module-structure on the associated sheaf simply by sheafifying the action morphism.
Let $$\mathcal{F}$$ be a presheaf of $$\mathcal{O}$$-modules, and let $$K=\coprod_{P\in X}\mathcal{F}_P$$ be the disjoint union of the stalks of $$\mathcal{F}$$. Now let $$\mathcal{H}$$ be the $$\mathcal{O}$$-module given by $$\mathcal{H}(U)=\{f:U\to K\mid \forall P\in U,\ f(P)\in\mathcal{F}_P\}$$ for all open $$U\subseteq X$$, with restriction maps as in the usual sense. Then we identify $$\mathcal{F}$$ with a subpresheaf $$\mathcal{G}$$ of $$\mathcal{H}$$. In particular, we take $$\mathcal{G}(U)=\{f:U\to K \mid \mbox{\exists s\in \mathcal{F}(U) such that, \forall P\in U, f(P)=s_P}\}$$ for all open $$U\subseteq X$$; $$s_P$$ is the germ of $$s$$ at $$P$$. The sheafication of $$\mathcal{G}$$ is given by \begin{align*}\mathcal{G}^+(U) = \{f:U\to K \mid \mbox{\forall P\in U, \exists V\in N_U(P) and t\in\mathcal{F}(V) such that, \forall Q\in V, f(Q)=t_Q}\}\end{align*} for all open $$U\subseteq X$$; $$N_U(P)$$ is the set of all open neighbourhoods of $$P$$ in $$U$$. Since $$\mathcal{G}^+$$ is a subsheaf of $$\mathcal{H}$$, it suffices to show that $$\mathcal{G}^+(U)$$ is an $$\mathcal{O}(U)$$-submodule of $$\mathcal{H}(U)$$ for all open $$U\subseteq X$$. Fix an open $$U\subseteq X$$. Observe that the $$0$$ function is in $$\mathcal{G}^+(U)$$. Now let $$r\in\mathcal{O}(U)$$, let $$f$$ and $$g$$ be in $$\mathcal{G}^+(U)$$, and fix $$P\in U$$. Then there exist $$V$$ and $$W$$ in $$N_U(P)$$, $$s\in\mathcal{F}(V)$$ and $$t\in\mathcal{F}(W)$$ such that $$f(Q)=s_Q$$ for all $$Q\in V$$, and $$g(R)=t_R$$ for all $$R\in W$$. Therefore, $$(rf+g)(Q)=rs_Q+t_Q=(rs+t)_Q$$ for all $$Q\in V\cap W$$, and thus $$rf+g\in \mathcal{G}^+(U)$$. Hence, $$\mathcal{F}^+\simeq\mathcal{G}^+$$ is an $$\mathcal{O}$$-module.