My interpretation of the question is "Do there exist non-isomorphic non-abelian groups $G$ and $H$ such that $|G|=|H|$ and they have the same number of elements of the same order". This is not the classification question, but I suspect this is the question the OP wants to know the answer to (and is asked in the last line).
Note that, as leo points out in the comments, the abelian case is covered elsewhere.
The solution to the non-abelian case is, perhaps, quite easy. As the OP points out, there exist abelian and non-abelian groups which have the same number of elements of any order, call them $A$ and $B$. So $A$ is abelian, $B$ is non-abelian, $|A|=|B|$ and we have the condition on the order of elements. The idea is simply to take $G=A\times B$ and $H=B\times B$.
However, cross-produts can introduce elements of new orders (for example, $\mathbb{Z}_2\times\mathbb{Z}_3\cong\mathbb{Z}_6$), so we have to be careful. I am not saying that the above construction doesn't always work, but rather you would have to prove that it always works. In order to get round this proof, take $B$ to be the following group.
$$B=\langle x,y,z; x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx\rangle$$
This group has order 27, exponent three and is non-abelian. To see this you should check that each of $(yz)^3$, $(y^2z)^3$ and $(yx^2)^3$ define the trivial element. Then take $A=\mathbb{Z}_3^3$ to be the abelian group of order 27 and exponent three. Taking $G=A\times B$ and $H=B\times B$ solve the problem!
EDIT You can find non-abelian groups of order $p^n$ and exponent $p$, $p>2$, by considering the subgroup $S_n^p$ of $GL_n(\mathbb{Z}_p)$ consisting of upper-triangular matrices, so $S_3^p$ consists of matrices of the following form.
$$\left(
\begin{array}{ccc}
1&\ast&\ast\\
0&1&\ast\\
0&0&1
\end{array}
\right)$$
The $3\times 3$ matrix groups constructed this way are called Heisenberg groups, and its isomorphism class is that of the extra-special $p$-group of exponent $p$. This means that you can find lots of non-isomorphic groups with the same orders and spectra.
Here's how I would look at it. Suppose you are given the order $g$ of the group which has canonical factorization $p_1^{\alpha_1} p_2^{\alpha_n} \dots p_n^{\alpha_n}$. By the fundamental theorem of finite group your group is direct product of $n$ abelian $p$-groups, which are in turn direct products of cyclic subgroups of different orders. We want to determine for each prime which $p$-group is being used.
To do this look at the elements whose order has only $p$ as a prime divisor. Those are the elements which are of the form $(1,1,1,g,1,1\dots 1)$ These elements form a subgroup which is isomorphic to the $p$-subgroup.
So we would like to discern which subgroup it is by knowing the elements of order power of $p$.
To prove we can do this we need to prove that if
$\mathbb Z_{p^{a_1}}\times\mathbb Z_{p^{a_2}}\dots \mathbb Z_{p^{a_r}}$ and $\mathbb Z_{p^{b_1}}\times\mathbb Z_{p^{b_2}}\dots \mathbb Z_{p^{b_s}}$ are groups of order $p^n$ with different exponents they have a different count of orders.
To see this note that the order of an element in a direct product $g=(g_1,g_2\dots g_n)$ is the least common multiple of all of the orders, in a $p$-group it is the highest order.
So go order the factors from least to greatest. and suppose they differ for the first time at factor $k$, where the first group has a larger exponent (call it $p^x$) Then that group has more elements of order $p^x$.
Best Answer
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.