= Solution
Let $|g|_S$ be the least length of a word in $S\cup S^{-1}$ representing $g$, and write $B_S(r)=\{g:|g|_S\le r\}$. <Polynomial growth of a group> of degree at most $d$ means that, for some constant $C$,
$$
|B_S(r)|\le C(1+r)^d\qquad(r\ge0).
$$
In particular these balls are finite. For the usual convention $S$ is finite; finiteness of $B_S(1)$ would force that in any case.
For a finite subset $T$, set $L=\max(1,\max_{t\in T}|t|_S)$. Replacing each letter of a $T$-word by an $S$-word gives
$$
B_T(r)\subseteq B_S(Lr).
$$
Thus $|B_T(r)|\le C(1+Lr)^d\le CL^d(1+r)^d$. This proves the same degree bound for $\langle T\rangle$. If $T$ is empty the <subgroup> is trivial and the assertion is immediate.
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