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 $S_3,\mathbb{Z}_6$.
You need no contradiction, but you're missing a couple of points.
Let $gH$ be an element of finite order in $G/H$. Therefore there exists $n>0$ with $(gH)^n=H$, that is, $g^nH=H$. Hence $g^n\in H$, so there is $m>0$ with $(g^n)^m=1$. As a consequence $g^{nm}=1$, so $g$ has finite order, hence $g\in H$ and $gH=H$.
Therefore the only element of finite order of $G/H$ is the identity $H$.
The points you are missing: $g^n\in H$ doesn't mean by itself that $g\in H$; there is no assumption that $H\ne G$.
Best Answer
Since $G$ is Abelian, I’ll write it additively. Say that $A\subseteq G$ is independent if $\sum F\ne 0_G$ for all finite $F\subseteq A$. Use Zorn’s lemma to get a maximal independent $B\subseteq G$. Show that for each $x\in G$ there is a finite $F_x\subseteq B$ such that $x=\sum F_x$; then show that $F_x$ is unique and infer that $G$ is a vector space over $\Bbb F_2$ with basis $B$.
Now modify the construction by first getting a basis $B_H$ for $H$ and then extending it to a basis $B$ for $G$, and show that $G/H$ is isomorphic to the subgroup generated by $B\setminus B_H$.