= Solution
The assertion is false. Let
$$
R=k[x,y],\qquad S=\{1,y,y^2,\ldots\},
\qquad A=k[x,y,x/y]\subseteq S^{-1}R.
$$
The element $1/y$ does not belong to $A$: assigning degrees $\deg x=\deg y=1$ and $\deg(x/y)=0$ shows that every element of $A$ has nonnegative total degree, whereas $1/y$ has degree $-1$.
If $A=T^{-1}R$ for some multiplicative subset $T$, the membership $x/y\in T^{-1}R$ would give $x/y=a/t$ for some $a\in R$ and $t\in T$. Since $R$ is a <unique factorization domain> and $x,y$ are coprime, the equation $xt=ay$ implies $y\mid t$. Write $t=yu$. As $t$ is invertible in $T^{-1}R$, so is
$$
\frac1y=\frac u t,
$$
contrary to $1/y\notin A$. Thus a subring of a <localization of a ring> need not itself be a localization of the original ring.
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