Assuming $H$ is a non normal subgroup of order $2$.
Consider Action of $G$ on set of left cosets of $H$ by left multiplication.
let $\{g_iH :1\leq i\leq 3\}$ be cosets of $H$ in $G$.
(please convince yourself that there will be three distinct cosets)
we now consider the action $\eta : G\times\{g_iH :1\leq i\leq 3\} \rightarrow \{g_iH :1\leq i\leq 3\}$ by left multiplication.
i.e., take an element $g\in G$ and consider $g.g_iH$ as there are only three distinct sets we get $g.g_iH = g_jH$ for some $j\in \{1,2,3\}$
In this manner, the elements of $g\in G$ takes cosets $g_iH$ (represented by $i$) to cosets $g_jH$ (represented by $j$)
i.e., we have map $\eta :\{1,2,3\} \rightarrow \{1,2,3\}$
which can be seen as $\eta : G\rightarrow S_3$
we know that $Ker(\eta)$ is normal in $G$ which is contained in $H$.
As $H$ is not normal in $G$ we end up with the case that $Ker(\eta)=(1)$ i.e., $\eta$ is injective.
i.e., we have $G$ as a subgroup(isomorphic copy) of $S_3$. But, $|G|=|S_3|=6$. Thus, $G\cong S_3$.
So, for any non abelian group $G$ of order $6$ we have $G\cong S_3$.
For an abelian group of order $6$ we have already know that $G$ is cyclic and $G\cong \mathbb{Z}_6$.
So, only non isomorphic groups of order $6$ are $S_3,\mathbb{Z}_6$.
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
If $q\mid p-1$ then $\rm{Aut}(C_p)$ has a unique subgroup of order $q$, and the map embedding $C_q$ in $\rm{Aut}(C_p)$ gives a semidirect product, which is not abelian (easy to check).
On the other hand, if $G$ is some other non-abelian group of order $pq$ then it is an easy exercise that $G$ has a normal subgroup of order $p$ and since $G$ also has a subgroup of order $q$, it must be a semidirect product. But since the subgroup of order $q$ in $\rm{Aut}(C_p)$ was unique, the only possibility is the one we already accounted for.