The answer is no in general. It is yes if $R$ is $\mathfrak{m}$-adically separated (e.g. noetherian), where $\mathfrak{m}$ is the maximal ideal.

For a counterexample, assume $R$ is a valuation ring (not a field) and $\mathfrak{m}=\mathfrak{m}^2$. Take $F=\mathfrak{m}$, $e=$ the inclusion in $R$. Now $e\otimes \mathfrak{m}$ is injective by flatness (over a valuation ring, flat=torsion free) and surjective because its image is $\mathfrak{m}^2$.

Now assume $R$ separated, let $k$ be the residue field, and put $\overline{F}=F\otimes k$. Then $e\otimes k$ is a linear form $\overline{F}\to k$ inducing an isomorphism $\overline{F}\otimes\overline{F}\to \overline{F}$, which immediately implies $\overline{F}=0$ or $\overline{F}=k$ (and $e\otimes k$ is an isomorphism in the second case).

If $\overline{F}=0$, then $F=\mathfrak{m}F$, hence $F=\mathfrak{m}^nF$ for all $n$, hence $e(F)\subset \mathfrak{m}^n$ for all $n$. By our assumption this implies $e=0$, and finally $F=0$.

In the other case, $e$ is surjective (its image is not contained in $\mathfrak{m}$), hence $F= R\oplus\ker(e)$. Tensoring with $F$ we get that $F\otimes \ker(e)$ must be zero, hence $\ker(e)=0$ since $R$ is a direct summand of $F$. Hence $e$ is an isomorphism. (In this case the assumption on $R$ is not used).