Is the smash product of two Moore spaces again a Moore space? Write $M(G,n)$ for the Moore space with $\tilde{H}_\ast(M(G,n);\mathbb{Z})$ naturally isomorphic to $G$ concentrated in degree $n$. 
Now fix finitely generated (Abelian) groups $G$ and $H$ and natural numbers $n$ and $m$. Under what conditions is the smash product $M(G,n)\wedge M(H,n)$ homotopy equivalent to a Moore space? 
By the Künneth theorem this is clearly the case when one of $G$ or $H$ is a free Abelian group. 
 A: The (reduced) Künneth theorem applied to $X = M(G,n) \wedge M(H,n)$ tells you that:
$$\tilde{H}_i(X) = \begin{cases}
G \otimes_{\mathbb{Z}} H & i = 2n \\
\operatorname{Tor}_1^{\mathbb{Z}}(G,H) & i = 2n+1 \\
0 & i \neq 2n, 2n+1
\end{cases}$$
Clearly if $\operatorname{Tor}(G,H) = 0$ then $M(G,n) \wedge M(H,n) \simeq M(G \otimes H, 2n)$. If $\operatorname{Tor}(G,H) \neq 0$, then the only possibility is when $G \otimes H = 0$ (in which case it's a $M(\operatorname{Tor}(G,H), 2n+1)$ space), and otherwise it's not a Moore space.
Since $G$ and $H$ are finitely generated, write:
$$G = \mathbb{Z}^r \oplus \bigoplus_i \mathbb{Z}/p_i^{a_i}\mathbb{Z},
\quad
H = \mathbb{Z}^s \oplus \bigoplus_i \mathbb{Z}/p_i^{b_j}\mathbb{Z}$$
(where the $p_i$ are prime numbers and $a_i, b_i \ge 0$). Assume that both $G \neq 0$ and $H \neq 0$ (otherwise it's obvious).
Clearly if $r > 0$ or $s > 0$, then $G \otimes H \neq 0$. So in this case you must have $\operatorname{Tor}(G,H) = 0$ for it to be a Moore space. Since $\operatorname{Tor}(\mathbb{Z}, \mathbb{Z}/q\mathbb{Z}) = 0$ and $\operatorname{Tor}(\mathbb{Z}/q\mathbb{Z}, \mathbb{Z}/q'\mathbb{Z}) \neq 0$ iff $q$ and $q'$ are not coprime (where $q$, $q'$ are powers of primes), so you want $p_i = p_j \implies a_i b_j = 0$ (i.e. for $p$ prime, the $p$-torsion of $G$ and the $p$-torsion of $H$ cannot be both nonzero).
Similarly, if there is some $i$, $j$ with $p_i = p_j$ and $a_i b_j > 0$, then $G \otimes H \supset \mathbb{Z}/p_i^{a_i}\mathbb{Z} \otimes \mathbb{Z}/p_j^{a_j}\mathbb{Z} \neq 0$. But then you have $\operatorname{Tor}(G,H) \supset \operatorname{Tor}(\mathbb{Z}/p_i^{a_i}\mathbb{Z}, \mathbb{Z}/p_j^{b_j}) \neq 0$ and so $X$ cannot be a Moore space.
This leaves us the case $G \otimes H = 0$, which happens when $r = s = 0$ and $p_i = p_j \implies a_i b_j = 0$. But then we saw that this implies $\operatorname{Tor}(G,H) = 0$ too, and then $X$ is acyclic (in particular it's a Moore space...).
Conclusion. For finitely generated abelian groups $G$ and $H$, $M(G,n) \wedge M(H,n)$ is a Moore space if and only if for all primes $p$, the $p$-torsion of $G$ and the $p$-torsion of $H$ cannot both be nonzero.
This includes the case you mentioned when $G$ and $H$ are free abelian, because then they have no $p$-torsion for any $p$.
Remark. There was nothing special about the fact that the same $n$ was on both sides of the smash product. This equally applies to $M(G,n) \wedge M(H,m)$, and this will become either a $(n+m)$ or a $(n+m+1)$ Moore space.
