If $F$ is any field, there is a subgroup of $\text{PGL}_2(F)$ given by fractional linear transformation of the form $z \mapsto az + b$ where $a \in F^{\times}, b \in F$. This is precisely the subgroup fixing the point at infinity in $\mathbb{P}^1(F)$. Taking $F$ to be a finite field $\mathbb{F}_q$ we obtain a family of (usually) nonabelian Frobenius groups of order $q(q - 1)$.
Now, we can further restrict $a$ to lie in any subgroup of $F^{\times}$; in particular, for $q$ odd, taking it to lie in the subgroup of quadratic residues gives a family of (usually) nonabelian groups of order $\frac{q(q-1)}{2}$. Taking $q = 7$ gives the desired group.
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 $S_3,\mathbb{Z}_6$.
Best Answer
If $G'$ has order less than $p^3$, then it is abelian. So we may assume that $G'$ has order exactly $p^3$, in which case $G/G'$ is of order $p^2$, and thus $2$-generated. let $x$ and $y$ be the two generators. Moreover, $G_3=[G,G']$ has order at most $p^2$, $G_4$ has order at most $p$, and so $G_5$ will certainly be trivial; that is, $G$ has class at most $4$.
Since $G$ is a finite $p$-group, $G'$ is generated by the basic commutators in $x$ and $y$. We know that $G_n/G_{n+1}$ is generated by the basic commutators of weight $n$. So $G_2/G_3$ is cyclic, generated by $[y,x]$; $G_3/G_4$ is generated by $[y,x,x]$ and $[y,x,y]$. And $G_4/G_5 = G_4$ is generated by the basic commutators of weight $4$; but the only basic commutators of weight $4$ that are nontrivial are $[y,x,x,x]$, $[y,x,x,y]$, and $[y,x,y,y]$, since there is no nontrivial basic commutator obtained as $[c,c']$ with $c$ and $c'$ of weight $2$.
Thus, $G'$ is generated by $[y,x]$, $[y,x,x]$, $[y,x,y]$, $[y,x,x,x]$, $[y,x,x,y]$, and $[y,x,y,y]$. Since $G'$ is of class at most $4$, they commute pairwise (the only commutator not of weight greater than $4$ is $[[y,x],[y,x]]$, which is trivial).