The original statement is:
If $ab = ac$ and $a \neq 0$, then $b = c$. (In particular, this shows that the number 1 of Axiom 4 is unique.)
By Axiom 6, we know that exists a real number y such that $y*a = 1$. Since multiplications are uniquely defined, we know that $y*(a*b) = y*(a*c)$. By the Axiom 2, we can rewrite it as $(y*a)*b = (y*a)*c$ but we do know that $y*a=1$ so we have $1*b = 1*c$. By Axiom 4, we know that $1*b = b$ and $1*c =c$ so we have $b = c$
Axiom 2: $x + (y + z) = (x + y) + z$ and $x(yz) = (xy)z$.
Axiom 4: There exist two distinct real numbers, which we denote by 0 and 1, such that for every real x we have $x + 0 = x$ and $1*x = x$.
Axiom 6: For every real number $x \neq 0$ there is a real number y such that $x*y = 1$.
Your proof looks fine. :)
I'm posting this CW answer so that users who confidently concur have something to vote on,
and so this question doesn't stagnate in the Unanswered Questions Queue.