Such subrings $\,R\subseteq \Bbb Q$ are characterized by the subset of $S$ of primes that become invertible in $\,R,\,$ i.e. $\,R = \Bbb Z[1/p\ :\ p\in S]$ is obtained by adjoing the inverses of all primes in $S$. One easily checks that this is a ring since fractions having denominators being products of such primes are closed under addition and multiplication.
Conversely, let $\,R\,$ be such a proper subring $\,\Bbb Z\subsetneq R\subset \Bbb Q,\,$ and let $\,r = a/b\in R\,$ be a noninteger. Wlog $\,(a,b) = 1\,$ so by Bezout $\,aj+bk = 1\,$ for some $\,j,k\in\Bbb Z.\,$ Thus $\,j(a/b) + k = 1/b\in R.\,$ Hence $\,a/b\in R\iff 1/b\in R,\,$ so $R$ is generated by adjoining inverses of integers to $\,\Bbb Z.\,$ However $n$ is invertible iff all its prime factors are invertible, so we can restrict to inverses of primes.
This generalizes to any Bezout domain, i.e. rings like $\,\Bbb Z\,$ where gcds enjoy a Bezout identity. See here for further discussion. Said in the language of commutative algebra, every overring (of fractions) of a Bezout domain is a localization, i.e. is generated by adjoining inverses of elements.