Is Kaplansky's theorem for hereditary rings a characterization?

To answer the second question, no, it is not a characterization.

For example, let $k$ be a field and let $A=k[x]/(x^2)$. Then $A$ is not hereditary, but every $A$-module is a direct sum of copies of $A$ and of $k=A/(x)\cong Ax$, both of which are ideals.

(As mentioned by rschwieb in comments, my claimed classification of $A$-modules follows from more general results. But there's a fairly simple direct proof. Let $M$ be an $A$-module. Choose a basis $\{n_i\}$ of $Mx$ together with a choice of elements $\{m_i\}$ such that $n_i=m_ix$. Now extend $\{n_i\}$ to a basis $\{n_i\}\cup\{k_j\}$ of the kernel of multiplication by $x$. Then $\{m_i\}\cup\{n_i\}\cup\{k_j\}$ is a basis of $M$, for each $i$ the elements $m_i$ and $n_i$ span a submodule isomorphic to $A$, and for each $j$ the element $k_j$ spans a submodule isomorphic to $k$.)