If $G$ is non-abelian group of order 6, it is isomorphic to $S_3$

Here is a hands-on method.

Note that $G$ has an element $a$ of order $3$ hence at least two as $a^2$ has order $3$, but can't have an element of order $6$ or it would be cyclic and hence abelian.

Suppose the elements of order $2$ are $b,c,d$, then the elements of the group are $1,a,a^2,b,c,d$. No element of order $3$ can commute with any element of order $2$ else the product would have order $6$

Now $ab\neq ba$ implies both $aba^{-1} \neq b$ and $bab^{-1} \neq a$ - so neither the elements of order $3$ nor those of order $2$ can have a trivial action.

If you Sylow Theorems, you get 1 Sylow 3-subgroup and either 1 or 3 Sylow 2-subgroup(s).

If G has 1 Sylow 2-subgroup, it must be normal since it is a unique subgroup of given order. So we can take direct product of these two subgroups, which is isomorphic to G. But since each Sylow subgroups here are isomorphic to $\mathbb{Z}_2$ and $\mathbb{Z}_3$, we have that $G \cong \mathbb{Z}_2 \times \mathbb{Z}_3$, which contradicts that $G$ is not abelian.

So G has 3 Sylow 2-subgroups. Now you can explicitly list out elements of $G$ to see why it is isomorphic to $S_3$