We want to prove by induction on $n$ that every subgroup of an abelian group generated by $n$ elements is finitely-generated. If $n=1$, it is clearly true. From now on, suppose it is true for $n$.
Let $G= \langle a_1, \dots, a_{n+1} \rangle$ be an abelian group, $H \leq G$ be a subgroup and $\rho : G \to G/ \langle a_{n+1} \rangle$ be the quotient map. Notice that the group $G/ \langle a_{n+1} \rangle$ is generated by $\{ \rho(a_1) , \ldots, \rho(a_n) \}$, so by our induction hypothesis, the subgroup $\overline{H}:= \rho(H)$ is finitely generated: let $X=\{h_1,\dots, h_m\} \subset H$ be such that $\rho(X)$ generates $\overline{H}$.
On the other hand $H \cap \langle a_{n+1} \rangle$ is a cyclic group, say generated by $h_{m+1} \in H$. Now, we want to prove that $Y= \{h_1, \dots, h_m,h_{m+1} \}$ generates $H$.
Let $h \in H$. Of course, there exists a word $w \in \langle h_1,\dots, h_m \rangle$ such that $\rho(w)=\rho(h)$. Therefore, $h=w+k$ for some $k \in \mathrm{ker}(\rho)= \langle a_{n+1} \rangle$. Furthermore, $k=h-w \in H$ so $k=p \cdot h_{m+1}$ for some $p \in \mathbb{Z}$ since $\langle a_{n+1} \rangle \cap H = \langle h_{m+1} \rangle$. Finally, $$h=w+p \cdot h_{m+1} \in \langle h_1, \ldots, h_m,h_{m+1} \rangle = \langle Y \rangle,$$
so $Y$ generates $H$.
Yes, as mentioned in the comments you can always take the quotient by the so-called commutator subgroup. That is, the group generated by all commutators $[x,y]=xyx^{-1}y^{-1}$ of elements of $G$.
This quotient is always abelian, and is referred to as the abelianization of $G$.
Now, the answer to your title question is that the quotient $G/H$ will be abelian iff $H$ contains the commutator, aka the "first derived subgroup": $[G,G]\le H$.
Best Answer
Let $G$ be a group.
If $G$ is a finite group in which every proper subgroup is abelian then $G$ must be a solvable group. Its proof is not so easy with elementary tools. Thus such groups must be meta-abelian groups. As far as I remember it is not true for infinite groups. (Paul Plummer supplied an example in comments.)
If $G$ is a finite solvable group in which every nontrvial quotient is abelian then $G'$ is a minimal normal subgroup. Thus, $G'$ is isomorphic to $\mathbb Z_p \times\mathbb Z_p \ldots \times \mathbb Z_p $. Hence $G$ is meta-abelian group. In that case, One can also observe that $Q\in Syl_q(G)$ for $p\neq q$, $Q$ is abelian. $S_3,A_4$ are examples of such groups.
But such groups need not be solvable in general. For example consider the group $G=(A_5\times A_5)\rtimes \mathbb Z_2$ where the action of the involution is the map $(a,b)\mapsto (b,a)$ on $A_5\times A_5$.
I think you should ask a more specific question though it seems that your question is too general.