We can describe a symtrivial module over a Dedekind domain $R$ with field of fractions $K$ as follows:
(1) Torsion-free symtrivial modules are submodules of $K$.
(2) The torsion part of a symtrivial module is symtrivial
(3) A symtrivial torsion module is a direct sum of symtrivial $p$-torsion modules
(4) A symtrival $p$-torsion module $M$ satisfies $M/pM=0$ or $R/pR$.
(5) If a $p$-torsion module $M$ satisfies $M/pM=0$ or $R/pR$, then it is the direct sum of a $p$-divisible $p$-torsion module and a cyclic module.
(6) If $0 => T => M => F => 0$ is an exact sequence, with $T$ torsion, $M$ symtrivial, and $F$ torsion-free, then for each prime $p$, either $F$ is $p$-divisible or $T$ has no $p$-torsion.
(7) If $T$ is torsion and symtrivial, $F$ is torsion-free and symtrivial, and for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-torsion, then $T \oplus F$ is symtrivial.
If none of my proofs are mistaken, the remaining questions are
(Q1) For $T$ torsion symtrivial, $F$ torsion-free symtrivial, such that for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-torsion, what is $Ext^1(T,F)$?
(Q2) Do all elements of $Ext^1(T,F)$ give symtrivial modules?
Proof
(1) This is immediate, by Todd Trimble's argument.
(2) Indeed, take a module $M$ with torsion submodule $T$ such that $a \otimes b \neq b \otimes a$ in $T\otimes T$, but $a\otimes b = b \otimes a$ in $M\otimes M$. The equality $a \otimes b = b\otimes a$ must be the consequence of finitely many relations, involving finitely many elements. Consider the submodule $M'$ generated by the whole torsion module and those finitely many elements. The torsion-free quotient of $M$'$ is a finitely generated submodule of $\mathbb K$, thus is a fractional ideal, thus projective, so the submodule splits into a direct sum of torsion and torsion-free parts. But than $T\otimes T$ is a direct summand of $M'\otimes M'$, so if $a\otimes b \neq b\otimes a$ in $T\otimes T$, they do not equal each other in $M'\otimes M'$ - but a complete set of relations implying that they do are relations of $M' \otimes M'$, a contradiction.
(3) This is immediate from things noted in the original question.
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.
Thus, a symtrivial module consists of the choice of a torsion symtrivial module, a torsion-free symtrivial module, and an appropriate extension between them. Next we determine which pairs of a symtrivial torsion and non-torsion group are possible.
The non-torsion part of $A$, localized at $p$, is either $\mathbb Z_p$ or $\mathbb Q_p$. If it is $\mathbb Z_p$ then the exact sequence splits and thus the $p$-torsion part must be trivial. If it is $\mathbb Q_p$, the $p$-torsion part can be any symtrivial $p$-torsion group. If you take the direct sum of any pair of a symtrivial torsion group and a symtrivial non-torsion group that satisfy the condition that, at every prime, either the torsion is trivial or the non-torsion is divisible, you get a symtrivial group, so there is always at least one example.
Thus all that remains is to find out what exts are possible. I'm not sure how to go about that.
Everything so far works exactly the same over a general Dedekind domain.