I believe, though a not sure, that any two ideals $A, B$ of a Dedekind domain $X$ are isomorphic as $X$-modules iff their localizations $A_p, B_p$ are isomorphic for any prime ideal $p$. Could anyone either prove this, or provide a counterexample?
This is as far as I've gotten:
Let $\phi_p$ be an isomorphism $A_p \to B_p$, then since $X$ is Dedekind there exists a $c\in Q(X)$ (where $Q(X)$ is the quotient field of $X$) such that, treated as a function $Q(x) \to Q(x)$, $\phi_p(x) = c_px$. Multiplying by a $s_p$ not in $p$ we obtain $s_p\phi_p(A)\subset B$. Now, clearly $s_p$ generate $X$, and so $X$ being finitely generated we can find finitely many $x_i \in X$ such that $\sum{x_i s_{p_i}} = 1$, for some suitable indexing. Set $\phi(x) = \sum{x_i s_{p_i} \phi_{p_i}(x)} = (\sum{x_i s_{p_i} c_{p_i}})x$. Then clearly $\phi$ is an injection into $B$, if it is non-zero.
However, nothing so far indicates that this function is surjective (or even non-zero).
Your claim is wrong (the non-trivial part, of course). If $R$ is a Dedekind domain, then $R_p$ is a DVR for any prime ideal $p$. In particular, any two non-zero ideals of $R_p$ are isomorphic (to $R_p$). Now you only have to consider a Dedekind domain which is not a PID and two non-zero ideals $I,J$ with $I$ principal and $J$ not, and here you are a counterexample.