Can one show that there must be only two groups of order 6 and that every abelian group of order 6 must be isomorphic to $C_6$ without doing an entire classification of groups (i.e. showing from the start that all groups of order 6 must be isomorphic to either $C_6$ or $S_3$)?
The context of this query is that I want to easily show that $\operatorname{GL}_2(\mathbf{F}_2)$ is isomorphic to $S_3$ without doing the whole work of classifying groups of order $6$ (and so the with the stategy above all that remains is to show that $\operatorname{GL}_2(\mathbf{F}_2)$ and $S_3$ are non-abelian and hence they must be isomorphic with the information above).
Suppose $G$ is any group with $|G| = 6$. There exist $a\in G$ of order $2$ and $b\in G$ of order $3$ by Cauchy's theorem. If $G$ is abelian, then $ab$ has order $6$, and so $G \cong C_6$. Otherwise, the subgroup $N = \langle b \rangle$ is normal because it has index $2$, and $N$ and $H = \langle a \rangle$ have trivial intersection and satisfy $NH = G$, so $G$ is a semidirect product of $N$ and $H$, which is defined by a nontrivial (since $G$ is not abelian) homomorphism $\phi: H \to \text{Aut}(N) \cong C_2$, of which there is only one. So there are only two groups of order $6$, and only one is abelian, namely $C_6$.
Here's an alternative, more elementary approach. I use Lagrange's theorem, but that's certainly something you'll learn very soon if you haven't already. I'll show there's only one nonabelian group of order $6$, since that seems to be what you really need. (This ends up basically being a proof of the classification, so maybe it's not so great.)
Suppose $G$ is a non-abelian group with $|G| = 6$. By Lagrange's theorem, the order of any element of $G$ is $1$,$2$,$3$, or $6$. If there is an element of order $6$, then $G$ is cyclic and thus abelian, so there is no element of $G$ with order $6$. If every nonidentity element of $G$ has order $2$, then if $a,b\in G$, we have $abab = (ab)^2 = 1$, and thus, multiplying both sides on the right by $ba$, we have $ab = ba$, so $G$ is abelian, and thus there must be some element of $G$ with order $3$. If every nonidentity element has order $3$, then letting $a \in G$ have order $3$, let $b\in G \setminus \{1,a,a^2\}$, and so $b^2 \notin \{1,a,a^2\}$, since otherwise $(b^2)^2 = b$ would be in this set, and it is not. Now let $c \in G \setminus \{1,a,a^2,b,b^2\}$. By similar reasoning, $c^2 \notin \{1,a,a^2,b,b^2\}$, but also $c^2 \ne c$ since $c$ has order $3$. Therefore $G$ contains at least $7$ elements, which is not the case. So there must be some element of $G$ with order $2$.
Let $a\in G$ have order $2$ and $b\in G$ have order $3$ (possible by the above paragraph). You can verify easily that $G = \{1, a, b, b^2, ab, ab^2\}$, as any equality between two of these elements leads quickly to a contradiction (of the facts that $a$ has order $2$ and $b,b^2$ have order $3$), and there are six of them. You can also check that $aba$ is equal to either $b$ or $b^2$ by eliminating the other possibilities in a similar way (equate $aba$ to the other elements in $\{1, a, b, b^2, ab, ab^2\}$ and reach contradictions easily). If $aba = b$, then by multiplying on the right by $a$, we get $ab = ba$, and so $G$ is abelian since $a$ and $b$ generate $G$ and commute. Therefore $aba = b^2$, which implies $ba = ab^2$. You can use this rule to reduce any string of $a$s and $b$s to the form $a^ib^j$ with $0\le i \le 1$ and $0 \le j \le 2$, which is the form in our list $G = \{1, a, b, b^2, ab, ab^2\}$, and so this rule determines the multiplication rule on $G$. Since $a = (12)$ and $b=(123)$ in $S_3$ satisfy these rules, $G$ is isomorphic to $S_3$.