I love this question! I've enjoyed thinking of it. 

By naturality, the sequence splits always if and only if it splits for $X=K(A,1)$ with $A$ an abelian group ($A=H_1(X,\mathbb{Z})$). Actually, the sequence splits for $X=K(A,1)$ if and only if it splits for all spaces with $H_1(X,\mathbb{Z})=A$.  In case $X=K(A,1)$, we have an isomorphism of short exact sequences:
$$
\begin{array}{cccccccc}
0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\
\downarrow&&\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\
0&\to& H_2(X,\mathbb{Z}/2)&\to& \pi_2^{st}(X)&\to&H_2(X,\mathbb{Z})&\to&0
\end{array}
$$
Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. See:

Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub

Now the problem is reduced to a purely algebraic one. It is easy to see that the top sequence splits in many cases, e.g.~if $A$ is finitely generated, using the structure theorem. Actually, $A\hat{\otimes}A$ and $A\wedge A$ are quadratic functors on $A$, both with cross effect $\otimes$, the tensor product. Moreover, the quotient natural map $A\hat{\otimes}A\to A\wedge A$ induces an isomorphism between cross effects. Therefore, the class of abelian groups for which the sequence splits is closed under direct sum. This class includes cyclic groups, $2$-divisible groups, and the Pruffer group $\mathbb{Z}/2^\infty$ (the 2-primary component of $\mathbb{Q}/\mathbb{Z})$. In all the previous cases, one of the terms of the short exact sequence vanishes. Hence, the class also includes, finitely generated groups, torsion groups, etc.

Unfortunately my knowledge of infinite abelian groups is not sufficient to find a counterexample (or a complete proof).

PS. What happens for $A=\mathbb{Z}_2$ the $2$-adic integers?