Let $z$ be a word, and let $M=FM(a,b)/z^{-1}$. If we wish $a$ and $b$ to be invertible in $M$, then $z$ must contain both $a$ and $b$ at least once. (For example, $FM_2[a^{-n}]=FM_2[a^{-1}]\cong F(a)\ast FM(b)$ by the canonical maps.)

Suppose $z = a^nwa^m$ with $n, m>0$. If $z^{-1}$ exists, then $a$ has a right inverse $r = a^{n-1}wa^mz^{-1}$ and a left inverse $l = z^{-1}a^nwa^{m-1}$, and in fact these two are equal by the standard argument ($l = lar = r$); so just say $l=r=a^{-1}$. Then $z^{-1}a^nwa^m = e$ implies $z^{-1}a^nw = a^{-m}$ implies $a^mz^{-1}a^nw = e$ ($w$ has a left inverse), and similarly $w$ has a right inverse which is the same. If $z$ contains any occurrences of $b$, then we choose $w$ to begin and end with $b$; by the earlier argument applied to $w$, $b$ is invertible.

Symmetrically, if $z$ begins and ends with $b$ but contains an occurrence of $a$, then $z^{-1}$ exists implies $a^{-1}$, $b^{-1}$ exists.

I suspect these are the only cases which work (i.e., inverting a word of the form $awb$ or $bwa$ would not invert $a$ or $b$) but cannot yet prove it. However it is clear that inverting one of $a$ or $b$ will also invert the other (assuming $z$ contains at least one occurrence of each).

Let $z$ be a word, and let $M=FM(a,b)/z^{-1}$. If we wish $a$ and $b$ to be invertible in $M$, then $z$ must contain both $a$ and $b$ at least once. (For example, $FM_2[a^{-n}]=FM_2[a^{-1}]\cong F(a)\ast FM(b)$ by the canonical maps.)

Suppose $z = a^nwa^m$ with $n, m>0$. If $z^{-1}$ exists, then $a$ has a right inverse $r = a^{n-1}wa^mz^{-1}$ and a left inverse $l = z^{-1}a^nwa^{m-1}$, and in fact these two are equal by the standard argument ($l = lar = r$); so just say $l=r=a^{-1}$. Then $z^{-1}a^nwa^m = e$ implies $z^{-1}a^nw = a^{-m}$ implies $a^mz^{-1}a^nw = e$ ($w$ has a left inverse), and similarly $w$ has a right inverse which is the same. If $z$ contains any occurrences of $b$, then we choose $w$ to begin and end with $b$; by the earlier argument applied to $w$, $b$ is invertible.

Symmetrically, if $z$ begins and ends with $b$ but contains an occurrence of $a$, then $z^{-1}$ exists implies $a^{-1}$, $b^{-1}$ exists.

I suspect these are the only cases which work (i.e., inverting a word of the form $awb$ or $bwa$ would not invert $a$ or $b$) but cannot yet prove it. (EDIT: This is false, see Mark's comment below.)

EDIT: Partial result! Consider the structure $M$ whose underlying set is

$\{(m,e) \in \mathbb{N}\times\mathbb{Z} : m\ge e\}$

with binary operation $(m,e)\ast(m',e') = (\max(m,e+m'), e+e')$. It is routine to check that this is a monoid. Let $z=awb$ contain $i$ copies of $a$ and $j$ copies of $b$, and consider the morphism $f\colon FM(a,b)\to M$ with $f(a) = (0,-j)$ and $f(b) = (i,i)$.

From counting it is clear that $f(z) = (m,0)$ for some $m\in\mathbb{N}$. If $m=0$ (as it will be for any word of the form $a^ib^j$, and many other words besides), then $f$ extends to a morphism $FM(a,b)[z^{-1}]\to M$.

But $f(a)$ has no left inverse; if $(m,e)\ast (0,-j) = (0,0)$, then $e=j$ and $\max(m,j)=0$, which is impossible ($j>0$ by assumption). Thus $a^{-1}$ cannot be a member of $FM(a,b)[z^{-1}]$, i.e., inverting $z$ does not invert $a$.

(Intuition: For each word in $FM(a,b)$, start at zero and read the word from left to right. For each $a$, descend $j$ steps; for each $b$, ascend $i$ steps; and track both your peak and your current position. If reading $z$ never gets you to the positive numbers, then adjoining $z^{-1}$ does not get you $a^{-1}$ (or similarly $b^{-1}$).)