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.
Best Answer
The fact that $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z\not\cong\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ is implied by $\Bbb Z/4\Bbb Z\not\cong\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ (which is much easier to check: one of these groups has no element of order $>2$ while the other does) and related to the Fundamental Theorem of finite(ly generated) Abelian groups. Long story short: one can decompose (non-cyclic) abelian groups into direct products of cyclic groups of prime power order. However, splitting prime powers in this process produces non-isomorphic abelian groups.
As I am aware, this might be a bit over the top for this simple question. A simpler thing to observe is the following: $$\operatorname{ord}((a_1,\dots,a_n))=\operatorname{lcm}(\operatorname{ord}(a_1),\dots,\operatorname{ord}(a_n))$$ It is not hard to see that $\Bbb Z/4\Bbb Z$ has $2$ elements of order $4$, $1$ of order $2$ and $1$ of order $1$ while $\Bbb Z/2\Bbb Z$ has $1$ element of order $2$ and $1$ of order $1$. Combining these gives a much more efficient way of computing the number of elements of a given order.
The process always works in the sense that if you find different numbers of elements of a given order (dividing the group order) they cannot be isomorphic as isomorphisms preserve the order of elements. However, the converse is not necessarily true, that is there are groups of the same order with the same number of elements for every order which are not isomorphic (see here for a counterexample).
Let's consider the group $\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. Write $\Bbb Z/4\Bbb Z=\{0,1,2,3\}$ and $\Bbb Z/2\Bbb Z=\{0,1\}$. As $\operatorname{lcm}(n_1,n_2,n_3)\ge\max\{n_1,n_2,n_3\}$ we conclude that for $(a_1,a_2,a_3)$ having order $2$ we have $a_1\ne1,3$. This leaves $2\cdot2\cdot2=8$ possibilites. However, if $a_1=a_2=a_3=0$ then $\operatorname{ord}(a_1,a_2,a_3)=1<2$. But this is the only exception to be found. Hence, there are exactly $7$ elements of order $2$.
For $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z$ we are immediately down to at most $4$ possibilites ($3$ remain in the end). This shows that they cannot be isomorphic.