Is $\mathbb{Z}_2 \times \mathbb{Z}_4^\infty$ isomorphic to $\mathbb{Z}_4^\infty$?
My intuition says no, but I have not been able to find a proof. It is true that each group can be embedded in the other, but this would only necessarily make them isomorphic if they were finite.
It is also not the case that $\mathbb{Z}_2\times A$ is never isomorphic to $A$, since $A$ could be the group $\mathbb{Z}_2^\infty$. Obviously $\mathbb{Z}_4^\infty$ is not isomorphic to $\mathbb{Z}_2^\infty$, since the former contains elements of order $4$ while every element in the latter has order two, but it does show that the argument wouldn't be so simple.
Elements of order $2$ are $2$-divisible in the group $\mathbb{Z}_4^{\infty}$ but not in $\mathbb{Z}_2 \times \mathbb{Z}_4^{\infty}$ :
If $x \in \mathbb{Z}_4^{\infty}$ is an element of order $2$, there exists $y \in \mathbb{Z}_4^{\infty}$ such that $x = 2y$ (because all the components of $x$ are either $0$ or $2$).
There is no such $y$ if you take $x = (1, 0) \in \mathbb{Z}_2 \times \mathbb{Z}_4^{\infty}$.