Suppose $G, H$ are finite abelian groups with the same number of elements of any given order. Show they are isomorphic.
Since finite abelian groups are isomorphic if and only if their Sylow subgroups are, we may restrict our attention to the case where $G, H$, are $p$-groups, but I can't quite make it past there...
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.
Here's a "low-tech" approach, which yields an explicit formula of elementary divisors using the information of the number of elements of each order.
First we count the number of elements of each order in $\mathbb Z/p^n\mathbb Z$. If $m=d_n\cdots d_1$ is the $p$-adic expansion of $m\in\mathbb{Z}/p^n\mathbb{Z}$, then its order is equal to $p^{m+1-l}$, where $l$ is the smallest index for which $d_i\neq0$. So the number of elements in $\mathbb Z/p^n\mathbb Z$ whose order is at most $p^k$ is equal to $p^k$, where $k\leq n$.
Now let $G$ be an abelian $p$-group, and let $N_k$ denote the number of elements of order $p^k$ in $G$. The fundamental theorem of finite abelian group asserts that there is a unique sequence of nonnegative integers $\lambda_1,\cdots,\lambda_m$ such that $G\simeq\bigoplus_{i=1}^m(\mathbb Z/p^{i}\mathbb Z)^{\lambda_i}$. Since an element in $\bigoplus_{i=1}^m(\mathbb Z/p^{i}\mathbb Z)^{\lambda_i}$ has order at most $p^k$ if and only if each of its component has order at most $p^k$, it follows from the result in the preceding paragraph that $$ N_0+\cdots +N_k=(\prod_{i\leq k}(p^i)^{\lambda_i})(\prod_{i> k}(p^{k})^{\lambda_i}) $$ for $k\geq0$. Thus $$\begin{eqnarray} N_0+\cdots+N_{k+1}&=&(\prod_{i\leq k+1}(p^i)^{\lambda_i})(\prod_{i> k+1}(p^{k+1})^{\lambda_i})\\ &=&(\prod_{i\leq k}(p^i)^{\lambda_i})(\prod_{i> k}(p^{k})^{\lambda_i})p^{\lambda_{k+1}\ +\cdots+\lambda_m}\\ &=&(N_0+\cdots+N_k)p^{\lambda_{k+1}\ +\cdots+\lambda_m} \end{eqnarray}$$ for $0\leq k<m$. We can use the above system of equations the express $\lambda_i$ in terms of $N_j$, which was to be proved.
In what follows, you can first reduce to the case of $p$-groups as you intended to, but it is not necessary.
I owe essentially this proof to user:Marvoir, who wrote a solution on French Wikiversity (I slightly simplified it, hopefully not introducting any mistake).
Let $a_1\mid a_2\mid\dots\mid a_r$ be the invariant factors of $G$, so $a_i>1$ and $G\cong(\Bbb Z/a_1\Bbb Z)\times(\Bbb Z/a_1\Bbb Z)\times\dots\times(\Bbb Z/a_r\Bbb Z)$, and similarly, let $b_1\mid\dots\mid b_s$ be those of $H$.
For every natural number $k$, the number of elements $x\in G$ such that $kx=0$ is the sum, over all divisors $d\ge1$ of $k$, of the numbers of elements of order $d$ in $G$. On the other hand, it is equal to $\prod_{i=1}^r\gcd(k,a_i)$. Therefore (since $G,H$ have the same number of elements of any order $d$): $$\forall k\in\Bbb N\quad\prod_{i=1}^r\gcd(k,a_i)=\prod_{j=1}^s\gcd(k,b_j).$$
Assuming wlog $s\le r$, we shall derive by induction on $s$ that this condition implies $r=s$ and $\forall i\quad a_i=b_i$.
Base case $s=0$: then $r=0$, because if $r\ge1$ then $1=\prod_{i=1}^r\gcd(a_1,a_i)=a_1^r\ge a_1>1$, contradiction.
Induction step: assuming the result holds when $s=S-1$ for some $S\ge1$, let us prove it still holds when $s=S$. So, let $1<a_1\mid a_2\mid\dots\mid a_R$ and $1<b_1\mid b_2\mid\dots\mid b_S$, with $S\le R$, be such that $$\forall k\in\Bbb N\quad\prod_{i=1}^R\gcd(k,a_i)=\prod_{j=1}^S\gcd(k,b_j).$$ Then, $$a_1^R=\prod_{j=1}^S\gcd(a_1,b_j)\mid a_1^S$$ hence $R=S$, and then $$\forall j\quad\gcd(a_1,b_j)=a_1$$ so that $a_1\mid b_1$, and (similarly, by symmetry) $b_1\mid a_1$, so that by cancellation of the first term of both products, the main hypothesis rewrites $$\forall k\in\Bbb N\quad\prod_{i=\color{red}2}^R\gcd(k,a_i)=\prod_{j=\color{red}2}^S\gcd(k,b_j)$$ and by induction hypothesis, the result follows.
