Two finite abelian groups with the same number of elements of any order are isomorphic

You don't say how much structure you have proven for abelian groups, so I will not assume much. If you do know some structure theorems, please let us know. (E.g., there is a theorem that if $G$ is abelian, and $a$ is an element of $G$ of maximal order, then there is a subgroup $H$ of $G$ such that $G = H\oplus \langle a\rangle$; do you know that?) Anyway...

Let $G$ and $H$ be finite abelian $p$-groups that have the same number of elements of each order. We want to prove that $G$ and $H$ are isomorphic.

Let $p^n$ be the largest order of an element of $G$ (and of $H$). If $n=1$, then $G$ and $H$ are elementary abelian $p$-groups; so they are vector spaces over $\mathbf{F}_p$, and since they have the same number of elements, they are isomorphic (same dimension).

Assume the result holds for abelian groups whose largest orders are $p^k$, and let $G$ and $H$ be groups satisfying our hypothesis and in which the largest elements have oder $p^{k+1}$.

Show that $pG$ and $pH$ have the same number of elements of each order, and that the elements of largest orders have order $p^k$. Apply induction to conclude $pG\cong pH$. Now see if you can leverage that to get $G\cong H$. If you need more help with those steps, please ask through comments.