Given $p,q$, with $p<q$ and $p|(q-1)$, there exist a non-abelian group of order $pq$ and it is unique up to isomorphism:
If $G$ is a non-abelian group, $|G|=pq$, then subgroup $H$ of order $q$ is normal (by Sylow theorem); let $K$ be its subgroup of order $p$. Then $HK\leq G$ (since $H\triangleleft G$) and $HK=G$ (since $|KH|=|H|.|K|/|H\cap K|=pq=|G|$). So,
$H\triangleleft G$, $K\leq G$, $HK=G$, $H\cap K=(1)$ $\implies G\cong H \rtimes K$.
Now, to get all possible (non-isomorphic) semidirect products $H$ by $K$, we have to consider homomorphisms $\phi \colon K \rightarrow Aut(H)$.
Since $K\cong \mathbb{Z}/p$, $H\cong \mathbb{Z}/q$, we know that $Aut(H)\cong \mathbb{Z}/(q-1)$.
As $p|(q-1)$, and $Aut(H)$ is cyclic group of order $q-1$, it contains unique subgroup of order $p$. So for any two non-trivial homomorphisms $\phi, \psi \colon K\rightarrow Aut(H)$, we have $\phi(K)=\psi(K)$, and $K$ is cyclic. Also, these homomorphisms are injective, since $K$ is of prime order. (Note that trivial homomorphisms will give direct product of $H$ by $K$, so $G\cong H\times K$, abelian group of order $pq$).
Therefore by following theorem semidirect products of $H$ by $K$ w.r.t. $\phi$ and $\psi$ are isomorphic:
If $K$ is a cyclic group and $\phi,\psi\colon K\rightarrow Aut(H)$ are two injective homomorphisms such that $\phi(K)=\psi(K)$, then these two homomorphisms give isomorphic semi-direct products of $H$ by $K$ (Alperin and Bell- Groups and Representations).
Now it is clear the existance and uniqueness (up to isomorphism) of non-abelian groups of order $pq$.
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.
Best Answer
Think about it this way, suppose that you have that $G=A\rtimes_\varphi B$ where $A,B$ are abelian. You then have a short exact sequence $0\to A\to G\xrightarrow{\gamma} B\to 0$ and a backmap $B\xrightarrow{\psi}G$ such that $\gamma\circ\psi=1_B$. If you assume that $G$ is abelian then the splitting lemma for $\mathbb{Z}$-modules tells you that the sequence $0\to A\to G\to B\to0$ splits and so $G\cong A\oplus B$. Thus, if $A\rtimes_\varphi B$ is abelian, then $A\rtimes_\varphi B\cong A\oplus B$. But, it's easy to check that this is the case if and only if $\varphi$ is trivial. So, non-trivial semidirect products induce non-abelian groups.