The answer is no. There indeed exists a nonabelian subsemigroup of a group all of whose quotient groups are abelian.
Lemma. Let $G$ be a bi-ordered group and $G_+$ its subsemigroup of positive elements. Then every homomorphism $G_+\to L$, $L$ a group, uniquely extends to $G$.
Proposition. Under the assumptions of the lemma, suppose that every proper quotient group of $G$ is abelian. Then every quotient group of $G_+$ is abelian.
Proof of the proposition: let $L$ be a group and $G_+\to L$ be surjective. So its extension to $G$, given by the lemma, is surjective but not injective. So $L$ is abelian.
Example. Let $F$ be Thompson's group of the interval (piecewise affine increasing self-homeos of $[0,1]$ whose breakpoints are dyadic and slopes are integral powers of $2$.). Then every proper quotient of $F$ is abelian, and $F$ is bi-orderable (e.g. $F_+$ can be defined as the set of nonidentity elements of $F$ whose first nontrivial slope is $>1$).
Proof of the lemma. Let $f:G_+\to L$ be a homomorphism, $L$ a group.
Claim a: $f(g^{-1}hg)=f(g)^{-1}f(h)f(g)$ for all $g,h\in G_+$ [note that for this to make sense, we need bi-orderability, i.e., invariance of $G_+$ under conjugation]. Indeed, $f(h)f(g)=f(hg)=f(gg^{-1}hg)=f(g)f(g^{-1}hg)$ which proves the claim.
Claim b: if $g_1,h_1,g_2,h_2\in G_+$ with $g_1h_1^{-1}=g_2h_2^{-1}$ then $f(g_1)f(h_1)^{-1}=f(g_2)f(h_2)^{-1}$.
Indeed, write the assumption as $g_1 h_1^{-1}h_2h_1=g_2h_1 $. Then $$f(g_2)f(h_1)=f(g_2h_1)=f(g_1 h_1^{-1}h_2h_1)=f(g_1)f(h_1^{-1}h_2h_1)=f(g_1)f(h_1)^{-1}f(h_2)f(h_1),$$ which is what is desired (the last equality used Claim a).
Claim b makes it valid to define $f(gh^{-1})=f(g)f(h)^{-1}$ for $g,h\in G_+$, so $f$ is now a well-defined map $G\to L$.
Claim c: $f$ is a homomorphism $G\to L$. Indeed, for $g_1,h_1,g_2,h_2\in G_+$, we have $$f(g_1h_1^{-1}g_2h_2^{-1})=f(g_1h_1^{-1}g_2h_1(h_2h_1)^{-1})=f(g_1h_1^{-1}g_2h_1)f(h_2h_1)^{-1}$$ $$=f(g_1)f(h_1^{-1}g_2h_1)f(h_2h_1)^{-1}=f(g_1)f(h_1)^{-1}f(g_2)f(h_1)f(h_1)^{-1}f(h_2)^{-1}$$ $$=f(g_1h_1^{-1})f(g_2h_2^{-1}).$$ (Claim a is used in the 4th equality.) So the lemma is proved.