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).