Let $S$ be an arbitrary set (countable or uncountable). It is clear that the free abelian group generated by $S$ is isomorphic to the direct sum
$$\bigoplus_{s\in S}\mathbb{Z}.$$
Is the free group generated by $S$ then isomorphic to
$$\ast_{s\in S}\mathbb{Z}$$ where $\ast$ denotes the free product of groups?
Yes.
It's a general pattern based on the free-forgetful adjunction: the free functor preserves colimits, in particular coproducts, so since $S=\coprod_{s\in S}1$ where $1$ is a set with one element, then $$F(S)=\coprod_{s\in S}F(1)\,,$$ and the coproduct in the category of groups is free product and $F(1)=\Bbb Z$.