How can we show the above assertion if in case it is true (namely that if a group $G$ is abelian and $f:G \to H$ is an isomorphism then $H$ is abelian also)? In particular every group is isomorphic to a finite group (by Cayley's Theorem). If the group $G$ is finite with order $n$ then it is isomorphic to $S_n$. But $S_n$ is not abelian for $n>3$. What can we then conclude about the group $G$? In particular, is G abelian or not?
You are misstating Cayley's theorem. It says that if $|G|=n$, then $G$ is isomorphic to a subgroup of $S_n$.
Now, when $\phi:G\to H$ is a homomorphism and $G$ is abelian, it follows that for $a,b\in G$, $$\phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a)$$ so $\phi(G)$ is an abelian subgroup of $H$. For example, the cyclic group $C_n$ embeds as the cyclic subgroup of $S_n$ generated by $(1\,2\,\ldots\,n)$.
I don't know what you are wanting to say about non-abelian groups.
Given a group $G$ of order $n$ you cannot decide in general whether or not $G$ is abelian. You need to know more about the group. For example, given a group of order $6$, this group may be abelian or may not be abelian. Both cases occur, namely $G=S_3$ and $G=C_6$.
In some special cases, however, you can "conclude". One case would be, that $n=p$ or $n=p^2$ with $p$ prime. Any group of that order must be abelian - see here.
A remark to your claim that any group $G$ of order $n$ must be isomorphic to $S_n$. This cannot be true because already the orders are different for $n\ge 3$.
To answer the question in the title:
Homomorphism images of abelian groups are always abelian.
Non abelian groups can have abelian homomorphic images: $G/G'$ is abelian, where $G'$ is the commutator or derived subgroup.