 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 $C$, 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 givesThus $|B_T(r)|\le C(1+Lr{)}^d\le C{L}^d(1+r{)}^d$. This proves the same degree bound for $⟨T⟩$. If $T$ is empty the subgroup is trivial and the assertion is immediate.