We can divide the classification, at least initially, into torsion and nontorsion parts.
Nontorsion we classify immediately. They are the submodules of $\mathbb Q$, by Todd's argument.
Torsion is more subtle. We can divide a torsion module into $p$-power torsion groups for each prime $p$. Let $A$ be a $p^n$-torsion group, then $A[p^n]/A[p^{n-1}]$ is an $\mathbb F_p$-vectors space that injects naturally into $A[p^{n-1}]/A[p^{n-2}]$. (Where $A[N]$ is the $N$-torsion part of $A$.) This gives a filtration of $\mathbb F_p$-vector spaces. We can choose a basis that agrees with this filtration, and by choosing lifts of the basis vectors we get an isomorphism between any two groups with the same filtration. In particular, we can write it as a direct sum of cyclic subgroups and the special module $\mathbb Q_p/\mathbb Z_p$.
The tensor of any two cyclic $p$-power group is nontrivial. The tensor of any cyclic group with $\mathbb Q_p/\mathbb Z_p$ is trivial. $\mathbb Q_p /\mathbb Z_p$ tensor itself is trivial.
So a torsion symtrivial group is a direct sum of, for each prime, up to $1$ $p$-power cyclic group and arbitrarily many copies of $\mathbb Q_p/\mathbb Z_p$.
Moreover, a symtrivial group must have a symtrivial torsion subgroup. Indeed, let $a \otimes b \neq b \otimes a$ in the torsion subgroup, but $a\otimes b = b \otimes a$ when the whole things is tensored. Then the proof that that inequality holds uses only finitely many elements of the group. Consider the subgroup generated by the whole torsion module and those finitely many elements. The nontorsion part of it is a finitely generated submodule of $\mathbb Q$, thus $\mathbb Z$, thus projective, so the torsion and nontorsion part split. This is a contradiction.
Thus, a symtrivial module consists of the choice of a torsion symtrivial module, a torsion-free symtrivial module, and an appropriate extension between them. Only the exts need classifying.