Here is an example: a product of infinitely many $\mathbb{RP}^\infty$'s.

The crucial thing thing to see is that $\mathbb{RP}^\infty$ (or, easier to see, its universal cover $S^\infty$) has a group structure whose underlying group is a vector space of dimension $2^{\aleph_0}$. This is not hard: the total space $S^\infty$ of the universal $\mathbb{Z}_2$-bundle is obtained by applying a composite of functors to the group structure $\mathbb{Z}_2$ in the category of sets:

$$Set \stackrel{K}{\to} Cat \stackrel{nerve}{\to} Set^{\Delta^{op}} \stackrel{R}{\to} Top$$

($Top$ here is the category of compactly generated Hausdorff spaces and continuous maps). Here $K$ is the right adjoint to the "underlying set of objects" functor; it takes a set to the category whose objects are the elements of the set and there is exactly one morphism between any two objects. The functor $R$ is of course geometric realization.

Each of these functors is product-preserving, and since the concept of group can be formulated in any category with finite products, a product-preserving functor will map a group object in the domain category to one in the codomain category. Even more: the concept of a $\mathbb{Z}_2$-vector space makes sense in any category with finite products since we merely need to add the equation $\forall_x x^2 = 1$ to the axioms for groups, which can be expressed by a simple commutative diagram.

Thus $S^\infty$ is an internal vector space over $\mathbb{Z}_2$ in $Top$. Its underlying group (in $Set$) is clearly a vector space of dimension $2^\{\aleph_0}$. We make take this vector space to be the countable product $\mathbb{Z}_2^{\mathbb{N}}$. Modding out by $\mathbb{Z}_2$ (modding out by a 1-dimensional subspace), the space $\mathbb{RP}^\infty$ is also, as an abstract group, isomorphic to this. And so is a countably infinite product $(\mathbb{RP}^\infty)^{\mathbb{N}}$ of copies of $\mathbb{RP}^\infty$.

Finally, the functor $\pi_1$ is product-preserving, and so

$$\pi_1((\mathbb{RP}^\infty)^{\mathbb{N}}) \cong \mathbb{Z}_{2}^{\mathbb{N}}$$

and we are done.

Here is an example: a product of infinitely many $\mathbb{RP}^\infty$'s.

The crucial thing thing to see is that $\mathbb{RP}^\infty$ (or, easier to see, its universal cover $S^\infty$) has a group structure whose underlying group is a vector space of dimension $2^{\aleph_0}$. This is not hard: the total space $S^\infty$ of the universal $\mathbb{Z}_2$-bundle is obtained by applying a composite of functors to the group structure $\mathbb{Z}_2$ in the category of sets:

$$\textbf{Set} \stackrel{K}{\to} \textbf{Cat} \stackrel{\text{nerve}}{\to} \textbf{Set}^{\Delta^{op}} \stackrel{R}{\to} \textbf{CGHaus}$$

($\textbf{CGHaus}$ here is the category of compactly generated Hausdorff spaces and continuous maps). Here $K$ is the right adjoint to the "underlying set of objects" functor; it takes a set to the category whose objects are the elements of the set and there is exactly one morphism between any two objects. The functor $R$ is of course geometric realization.

Each of these functors is product-preserving, and since the concept of group can be formulated in any category with finite products, a product-preserving functor will map a group object in the domain category to one in the codomain category. Even more: the concept of a $\mathbb{F}_2$-vector space makes sense in any category with finite products since we merely need to add the equation $\forall_x x^2 = 1$ to the axioms for groups, which can be expressed by a simple commutative diagram.

Thus $S^\infty$ is an internal vector space over $\mathbb{F}_2$ in $\textbf{CGHaus}$. It can also be considered an internal vector space over $\mathbb{F}_2$ in $\textbf{Top}$, the category of ordinary topological spaces, because a finite power $X^n$ in $\textbf{Top}$ of a CW-complex $X$ has the same topology as $X^n$ does in $\textbf{CGHaus}$ provided that $X$ has only countably many cells, which is certainly the case for $S^\infty$ (see Hatcher's book, Theorem A.6). Thus $S^\infty$ can be considered as an honest commutative topological group of exponent 2.

The underlying group of $S^\infty$ (in $\textbf{Set}$) is clearly a vector space of dimension $2^{\aleph_0}$. We make take this vector space to be the countable product $\mathbb{Z}_2^{\mathbb{N}}$. Modding out by $\mathbb{Z}_2$ (modding out by a 1-dimensional subspace), the space $\mathbb{RP}^\infty$ is also, as an abstract group, isomorphic to this. And so is a countably infinite product $(\mathbb{RP}^\infty)^{\mathbb{N}}$ of copies of $\mathbb{RP}^\infty$.

Finally, the functor $\pi_1$ is product-preserving, and so

$$\pi_1((\mathbb{RP}^\infty)^{\mathbb{N}}) \cong \mathbb{Z}_{2}^{\mathbb{N}}$$

and we are done.