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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 $am+bq=1$. am+bp=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$1/p\in R$. 1 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\$.