How to prove that $x = a^{-1} b$ is unique? ($a^{-1}$ is already proved unique.) In the text I read, it proved that $a^{-1}$ unique. But then it goes by saying uniqueness of $x = a^{-1} b$ follows from the fact that $a^{-1}$ is unique.

Update: Here is the complete text (from p.20 of Abstract Algebra by Dummit and Foote)

 A: The claim "The uniqueness of $x$ follows because $a^{-1}$ is unique" is incorrect and misleading. The proof uses only existence of $a^{-1}$ to deduce the uniqueness of solutions of $\,ax = b.\,$ Let's examine the proof closely. If $\,a,b,x\in G\,$ a monoid, and $\,a\,$ has a $\rm\color{#c00}{left}$ inverse $\,a'\in G,\,$ i.e.  $\,\color{#c00}{a'a = 1}\,$ then
$$\qquad\begin{align}  b \,&=\, ax\\
\Rightarrow\  a'b \,&=\, a'(ax)\\
&=\, (\color{#c00}{a'a})x \ \ \ {\rm by\ multiplication\ is\ associative}\\
&=\,  \color{#c00}1x\\
&=\, x\end{align}$$
Thus every root $\,x\,$ of $\,ax=b\,$ is equal to $\,a'b,\,$ so roots are unique - at most one root exists (since if $\,r_2,r_1$ are roots then $\,r_2 = a'b = r_1\Rightarrow\, r_2 = r_1).\,$  The proof did not use uniqueness of the (left) inverse $\,a'.\,$ Rather it used only its existence (to cancel $\,a).\,$ Even if - hypothetically - there were multiple inverses the proof would still work fine because it only needs to choose one of them.
The existence of a root follows if $\,a'\,$ is also a $\rm\color{#0a0}{right}$ inverse: $\,\color{#0a0}{aa'=1},\,$ since then the above arrow reverses by scaling by $\,a\,$ (or we can directly verify $\,x = a'b\,$ is a root: $\, ax = a(a'b) = (\color{#0a0}{aa'})b = b)$.
In the special case $\,b = 1\,$ the proof actually yields uniqueness of two-sided inverses, since it shows that if there exists a left inverse $\,a'$ of $\,a,\,$ i.e. $\,a'a = 1\,$ and there also exists a right inverse $\,x,\,$ i.e. $\,ax = 1,\,$ then $\,a' = x,\,$ i.e. left = right inverse, a result which is well-known. Thus - as above - if $\,a',a''$ are left inverses then $a'' = x = a'$ so $\,a'' = a'\,$ Similarly for uniqueness of right inverses.
More generally, to show that a set $S$ is a singleton it suffices to prove  $\,s\in S\iff s = a.\,$ Above is the special case when $\,S\,$ is a set of roots of an equation.
There is much further discussion of this subtlety in an old sci.math thread linked here (in the more concrete case of the additive group of real numbers). As is clear from that thread, this point often proves puzzling to beginners even in very concrete groups.
