In general, tensor products don't commute with direct products. As such, I understand that the natural map $(\prod\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Q} \to \prod (\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Q}) \cong \prod \mathbb{Q}$ is not an isomorphism since the denominators of $\prod \mathbb{Q}$ are unbounded. However, I am wondering if there is a noncanonical isomorphism of the spaces as $\mathbb{Z}$-modules.
My idea is the following: Define a $\mathbb{Q}$-module structure on $\mathbb{Z}^{\mathbb{N}} \otimes_\mathbb{Z} \mathbb{Q}$ by $q \cdot (p \otimes q') = p \otimes qq'$. Similarly, $q\cdot (q_1,q_2,...) = (qq_1,qq_2,...)$ defines a $\mathbb{Q}$-module structure on $\mathbb{Q}^{\mathbb{N}}$. Note that this $\mathbb{Q}$ action aligns with the $\mathbb{Z}$ action. It is easy enough to show that both $\mathbb{Z}^{\mathbb{N}} \otimes_\mathbb{Z} \mathbb{Q}$ and $\mathbb{Q}^{\mathbb{N}}$ are $\mathbb{Q}$-vector spaces of uncountable dimension, i.e., dim = $|\mathbb{R}|$. So by the axiom of choice, there exists an isomorphism $f$ between the two vector spaces. So automatically, $f$ is a bijection. Since the $\mathbb{Q}$ action aligns with the $\mathbb{Z}$ action, this means for all $m$, $f(zm) = zf(m)$ where we view the modules as $\mathbb{Z}$-modules. In other words, forgetting the $\frac{1}{n}$ action and only remembering $\mathbb{Z}$-action, we still have that $f$ is a $\mathbb{Z}$-linear map. And $f$ is still bijective as a set map. So then $f$ is an isomorphism of $\mathbb{Z}$-modules. Does this work?
I also understand that the $\mathbb{Q}$-basis of $\mathbb{Q}^\mathbb{N}$ is no longer a generating set of $\mathbb{Q}^\mathbb{N}$ as a $\mathbb{Z}$-module, but I only use the $\mathbb{Q}$-vector space structure to get that $f$ is a bijection which respects the $\mathbb{Z}$ action, so while $f$ is impossible to write down, at least it exists.