Let $S \subseteq T$ be two multiplicative sets in a commutative ring $A$. I need to show that
The homomorphism $\varphi: S^{-1}A \to T^{-1}A$, $ \ a/s \mapsto a/s$ is an isomorphism iff any element of $T$ is a factor of some element in $S$.
If any element of $T$ is a factor of some element in $S$ then clearly $\varphi$ is surjective. Also, if $\varphi(a/s) = a/s = a'/s' = \varphi(a'/s')$, then $\exists t\in T$ such that $t(as' - a's) = 0 $. But $dt = u \in S$ for some $d \in A$, so $u(as' -a's) = 0$ and hence $a/s = a'/s'$ in $S^{-1}A$. Therefore $\varphi$ is injective.
I am stuck with the proof for the other direction. If $\varphi$ is an isomorphism, then for any $x \in T \setminus S$ we have $\varphi(a/s) = a/s = 1/x$ for some $a \in A$ and $s \in S$. This is equivalent to $t(ax - s) = 0 \Leftrightarrow tax = ts$ for some $t \in T$. Can I deduce from this equality that $x$ divides some $s$ in $S$?
Since $t(ax - s) = 0$ we have $$\varphi(0/1) = 0/1 = t(ax - s)/1 = \varphi(t(ax - s)/1).$$ Thus $0/1 = t(ax - s)/1$ in $S^{-1}A$ by injectivity of $\varphi$ and it is equivalent to $ut(ax - s) = 0$ where $u \in S$. This implies the following equality: $$\varphi(u(ax - s)/1) = u(ax - s)/1 = ut(ax - s)/t = 0/t = 0/1 = \varphi(0/1).$$ Again, using the injectivity of $\varphi$, we obtain $u(ax - s)/1 = 0/1$ in $S^{-1}A \Leftrightarrow \ \exists u' \in S$ such that $u'u(ax-s) = 0$. Since $S$ is a multiplicative set $v = u'u \in S$ and $vax = vs$. Therefore $x | vs \in S$ and hence $x$ is a factor of an element in $S$.