Groups of order 6

Let $a\in G$. If $a\ne1,$ then $a$ has order $2,3,6$ (as you said). If $a$ has order $6$ then (as you said) $a^2$ has order $3$. If $a$ has order $2$ then we only have two elements in $G$. So let $b\in G$ (not equal to either $a$ or $1$) with order $2$. Note that if $(ab)^2=abab=1,$ then $a(abab)b=a1b\implies ba=ab.$ So we have only $\{1,a,b,ab\}$.

Now add $c$ (suppose it has order $2$ and is not already in $G$). We have at least $5$ elements now: $\{1,a,b,ab,c\}$. We want to show that there are at least $2$ more. Now if it has order $2,$ then $ac$=$ca$ (by the same argument for $ab$). Clearly $ac\ne 1,a,ab,c$. Suppose that $ac=b$. Then (since $a=a^{-1}$) $c=ab$ but $c\ne ab$. Similarly we can show that $bc$ is not equal to any element already in $G$. So $G$ contains at least $1,a,b,c,ab,ac,bc$.

If all nonidentity elements of a group have order $2$, then the group is Abelian. You can then show that any finite abelian group where every nonidenity element has order $2$ must have order a power of $2$.