I figured out how to prove it.
Let $(X, \mathscr{O})$ be a ringed space. Let $\mathscr{F}, \mathscr{G}$ be sheaves of $\mathscr{O}$-modules. Define $\mathscr{H}(U) = \mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$. Fix $p \in X$. Assume $U$ is an open n.h of $p$.
The $\mathscr{O}_p$-module structure on $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ induces a $\mathscr{O}(U)$-module structure on $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$ .
Define $\alpha_U : \mathscr{F}(U) \times \mathscr{G}(U) \to \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p, \quad (s,t) \mapsto s_p \otimes t_p$. This map is $2$-linear over $\mathscr{O}(U)$. Therefore $\alpha_U$ induces an $\mathscr{O}(U)$-module homomorphism from $\mathscr{F}(U) \otimes_{\mathscr{O}(U)} \mathscr{G}(U)$ to $\mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$. We shall abuse notation and also call this map $\alpha_U$.
Now forget about the $\mathscr{O}(U)$ module structure on the sections of $\mathscr{H}$. The $\alpha_U$s form a co-cone over the $\mathscr{H}(U)$s with $p \in U$ (that is they make the appropriate diagrams commute), therefore they induce a homomorphism of abelian groups $h : \mathscr{H}_p \to \mathscr{F}_p \otimes_{\mathscr{O}_p} \mathscr{G}_p$
Define $\psi : \mathscr{F}_p \times \mathscr{G}_p \to \mathscr{H}_p$ by $(s_p, t_p) \mapsto (s|_{U \cap V} \otimes t|_{U \cap V})_p$ where $s$ is a section of $\mathscr{F}$ over $U$ and $t$ is a section of $\mathscr{G}$ over $V$. It is easily verified that $\psi$ is 2-linear over $\mathscr{O}_p$. Therefore $\psi$ induces a $\mathscr{O}_p$ homomorphism from $\mathscr{F}_p \otimes_{p} \mathscr{G}_p$ to $\mathscr{H}_p$. It is easily verified that this map and $h$ are inverses.
If for example $\mathcal F = \mathcal O_X,$ then $\mathcal F(U)\otimes_{\mathcal O_X(U)}\mathcal G(U) = \mathcal O_X(U)\otimes_{\mathcal O_X(U)} \mathcal G(U) = \mathcal G(U)$, so in this case we see that the presheaf $\mathcal F\otimes \mathcal G$ is equal to $\mathcal G$, and so is a sheaf.
But this is not typical. A more illustrative example is $\mathcal F = \mathcal O(n)$ for $n > 0$ (on some positive dimensional projective space $\mathbb P^d$) and $\mathcal G = \mathcal O(-n) = \mathcal F^{-1}$. Then if $U = \mathbb P^d$,
we have $\mathcal G(\mathbb P^d) = 0$, and so the values of the presheaf
$\mathcal F\otimes\mathcal G$ on $\mathbb P^d$ is equal to $0$. On the other hand,
the actual (sheaf) tensor product $\mathcal F \otimes \mathcal G$ is equal to
the structure sheaf $\mathcal O_{\mathbb P^d}$ (because $\mathcal F$ and $\mathcal G$ are mutually inverse), which has a one-dimensional space of global sections.
So rather than trying to find situations when the presheaf tensor product is actually a sheaf, you will be better off getting an intuition for the sheafification process that goes into forming the sheaf tensor product from the presheaf version.
If $\mathcal F$ is a presheaf and $\overline{\mathcal F}$ the corresponding sheaf, then two key points are:
Thus, in the context of tensor products, there is always a canonical map $\mathcal F(U)\otimes \mathcal G(U) \to (\mathcal F\otimes\mathcal G)(U)$ (where the target denotes the sheaf tensor product), and the stalk of $\mathcal F\otimes\mathcal G$ at any point is the tensor product of the corresponding stalks of $\mathcal F$ and $\mathcal G$.
Finally, if $X =$ Spec $A$ is affine and $\mathcal F$ and $\mathcal G$ correspond to $A$-modules $M$ and $N$ respectively, then $\mathcal F\otimes\mathcal G$ is the quasi-coherent sheaf associated to the product
$M\otimes_A N$. In particular, if $U = $ Spec $A_f$ is a distinguished open
associated to $f \in A$, then the sections of $\mathcal F\otimes \mathcal G$
are equal to $(M\otimes_A N)_f = M_f\otimes_{A_f}N_f,$ which is the tensor product
of the sections of $\mathcal F$ and $\mathcal G$ on Spec $A_f$. So this is one case when one can work with the naive presheaf picture and get the correct answer, which is often useful in calculations (and also psychologically helpful in keeping a down-to-earth perspective on the general formalism).
Best Answer
No, in general this is not true.
Hint: Consider the case $\mathcal F=\mathcal G=\mathbb Z_X$ (the constant sheaf) on the discrete two point space $X=\mathrm{pt}\sqcup \mathrm{pt}$.