The answer is yes.  This follows from the fact that the group ${\mathbb Z}[S]$ should satisfy a universal property:

Let $A$ be any topological Abelian group and let $f: S \to A$ be a continuous function.  Then $f$ can be extended uniquely to a continuous homomorphism ${\mathbb Z}[S] \to A.$

Suppose $G$ and $H$ are two copies of ${\mathbb Z}[S]$ with possibly different topologies.  $G$ and $H$ both contain copies of $S.$  Applying the above unviersal property in two directions allows us to see that the identity from $G$ to $H$ is a homeomorphism.

[Edit: adding the existence portion of the argument]

Consider $(S \cup S^{-1} \cup \{1 \})^{\mathbb Z},$ where $S^{-1}$ is just another ``copy" of $S.$  An element $s \in S$ shall correspond to $s^{-1}$ in $S^{-1}.$  This is a topological space with the product topology.  Let $X$ be the subset consisting of all elements with only finitely many coordinates different from $1.$

Now we quotient $X$ in a technically complicated but conceptually simple way.  If $x \in X$ has $x_i = s$ and $x_{i+1} = s^{-1}$ for some $s \in S.$  Then we identify $x$ with $y$ where $y=x$ except at places $i,$ and $i+1$ where $y_i = 1 = y_{i+1}.$  We do the same if the $s,s^{-1}$ appear in the other order.
Also, if $\sigma$ is a permutation of ${\mathbb Z}$ then we identify $x$ with the $y$ where $y_i = x_{\sigma(i)}.$

To define a group operation, we take two infinite "words", slide them so that they have disjoint "support" and concatenate.  (Writing down the details here is harder than letting the reader guess what I mean.)