The answer is no in general ($e$ needs not to be in $\mathbb Z$), but one can show that $e$ is divisible in $R/\mathbb Z$ if and only if $e\in \mathbb Q\cap R$.
First let $R=\mathbb Z[1/p]$ for some prime number $p$. Then I claim that $1/p$ is divisible in $R/\mathbb Z$. Indeed for any $n\ge 1$, write $n=p^rm$ with $m$ prime to $p$. Let $a,b\in \mathbb Z$ such that $am+bq=1$. Then $$\frac{1}{p}=b+ \frac{am}{p}=b+n\frac{a}{p^{r+1}}\in \mathbb Z + nR.$$
For general $R$, denote by $D$ the elements $e\in R$ which are divisible in $R/\mathbb Z$. One can check directly that $D$ is a subring of $R$. Let us show $\mathbb Q\cap R\subseteq D$. If $e=k/q\in \mathbb Q\cap R$ with coprime $k, q$, then again using Bézout, $1/q\in R$. Then it is enough to show that $1/p\in D$ for all prime divisors $p$ of $q$. But this is done just above.
The converse is proved in Wilberd's answer ($e\in \mathbb Z[1/a]$).
Final remark: $\mathbb Q\cap R=\mathbb Z$ if and only if $\mathrm{Spec}(R)\to \mathrm{Spec}(\mathbb Z)$ is surjective. This is because the fiber of this morphism above $p$ is the spectrum of $R/pR$, and this spectrum is empty if and only if $1/p\in R$.