How the following step in the proof of this theorem is justified by group axioms? In my university, we have just did a proof of the following very basic theorem from linear algebra: "Let's have a group $(M,*)$ and $a,b\in M$. Then $x=a^{-1}*b$ is the only solution to the equation $a*x=b$."
I know how to show from the group axioms that $x=a^{-1}*b$ is a solution to the equation $a*x=b$.
However one thing is not clear to me when doing the second part of the proof, namely proving that $x=a^{-1}*b$ is the only solution to this equation. In school, we did the proof of this second part as follows:
"Suppose that $x'$ is also a solution to the equation $a*x=b$ . Therefore, $a*x'=b$. Than I can multiply both sides of the equation by $a^{-1}$ to get $a^{-1}*(a*x')=a^{-1}*b$. Then from association axiom I can move the backets as follows: $(a^{-1}*a)*x'=a^{-1}b$ and from another axiom I get $e*x'=a^{-1}b$ and from neutrality axiom I get $x'=a^{-1}b$."
The only step of this proof which I don't see to be justified by group axioms is the step where I'm basically saying that multiplying both sides of the equation by a same thing doesn't change the equality. It seems to be obvious, but shouldn't be there an axiom for that? Why not?

There is an axiom for that, namely:

$*$ is a binary operation on $G$. 

By definition, a binary operation on $G$ is a function 
$$G \times G \mapsto G
$$
which is denoted
$$(g,h) \mapsto g * h
$$
And since it is a function, it follows that if $(g,h) = (g',h')$ then $g * h = g' * h'$. Equivalently, if $g=g'$ and $h=h'$ then $g*h=g'*h'$.
So, using $g=g'=a^{-1}$, and using $h=a * x'$ and $h' = b$, and knowing by hypothesis that $h=a*x'=b=h'$, it follows that $g * h = g' * h'$, and then by substitution we get
$$a^{-1} * (a * x') = a^{-1} * b
$$
and you know where to go from there.
