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. More precisely, we get a map $G \to H$ and another $H \to G.$ Composing them one way gives us a map $G \to G$ which must be the identity by the universal property. Composing them the other way gives as a map $H \to H$ which again must be the identity. So the two maps are inverses and each 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 uniform 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.)