Proving From Spivak's Calculus Book, Basic Arithmetic Proof Background
Hello, I am reviewing calculus from a more challenging book, namely Michael Spivak's.  I find the exposition charming, but the proofs are challenging to a beginner like me.  I'm still trying to figure out what "moves" I'm allowed to use and which I am not.  And since he's building from the bottom up in the first chapter, I'm only allowed what he's given me so far. Thus my predicament.  My question is from the very first question in chapter 1.
The Problem
Prove the following:
If $ax = a$ for some number $a \ne 0$, then $x=1$.
My Proof
Proof 1. Suppose $ax = a$ for some number $a \ne 0$.  Multiplying both sides by $a^{-1}$ we get,
$$axa^{-1} = a^{-1}a$$
$$x(aa^{-1}) = 1$$
$$x = 1$$
Therefore, $x = 1$ when $ax = a$ for some $a \ne 0$. $\diamond$
Where I think I went wrong
I think this is incorrect because I think I've jumped to the conclusion I am trying to prove.  I can't multiply both sides by $a^{-1}$ because we are trying to establish that some nonzero number multiplying an unknown and yielding itself must mean that that unknown is 1. Am I correct here?
Here's my new attempt:
Proof 2.
Suppose $ax = a$ for some number $a \ne 0$.  Then adding $-a$ to both sides we get $ax + (-a) = a + (-a)$. By the additive inverse it follows that $ax + (-a) = 0$.  Distributing we get $a ( x - 1 ) = 0$.  Note that for every nonzero number $b$, there is a nonzero number $b^{-1}$ such that $b b^{-1} = b^{-1} b = 1$. This is the multiplicative inverse property, and a consequence of this is that if $ab = 0$, then either $a = 0$ or $b = 0$.  Returning to our equation we have $a ( x - 1 ) = 0$.  So by the multiplicative inverse property, we have either $a = 0$ or $(x - 1) = 0$. In this case we know that $a \ne 0$, since that was assumed in the antecedent.  So adding 1 to both sides we get $x + (- 1 + 1) = 1$.  This means $x = 1$ by the associative and additive inverse properties. $\diamond$
The Book's Proof
$1 = a^{-1}a=a^{-1}(ax)=(a^{-1}a)x = 1 \cdot x = x$
I have no idea what the book was doing.  My guess is it's using the multiplicative inverse property $aa^{-1} = a^{-1}a = 1$ to start.  Then it looks like he's substituting $a$ for $ax$ (he can do that because we assumed it correct?).  Then he's using associative and multiplicative inverse.  That's what it looks like to me.  I just don't know why he had to start with 1.  Why not start with the antecedents?
Thanks for your help.  Be gentle, I'm a newb, and have learned this on my own.
 A: First of all, congratulations, both on your motivation to relearn calculus like so, but also on your will to fully understand "minor" steps like these.
Your first proof was totally perfect. You didn't jump to any conclusions whatsoever. It is right because, firstly, you started with your hypothesis, namely, $ax=a$ for some $a \neq 0$. Then, you used one of the field axioms: the existence of multiplicative inverse for every real number different then zero. This implies that, because $a \neq 0$, there exists $a^{-1}$ and it is a real number. Multiplying both sides of your original equation by it keeps the equation right. Then, you used the field axioms of commutativity and multiplicative inverse to show that $x=1$.
Your second proof is also right. But it needs more detail on why $ab=0$ implies $a=0$ or $b=0$.
About Spivak's proof, I'll go step by step:
$1=a^{-1}a$ : 
Here, the author used the axiom of existence of multiplicative inverse for every $a \neq 0$.
$a^{-1}a =a^{-1}(ax)$: Then, the hypothesis was used. Because we are claiming that $ax=a$, the author substituted the factor of $a$ on the left hand side of the equation by $ax$, since $a=ax$.
$a^{-1}(ax) = (a^{-1}a)x$: This is just a simple use of the associative axiom for the real field.
$(a^{-1}a)x=1x$: Just the use of the definition of multiplicative inverse
$1x=x$: Again, follows from the fact (probably proven in the text) that $1x=x$, $\forall x$.
Best of luck.
A: I think André Muchon explained right.
The autor claims that for any number different from zero there is its multiplicative inverse such $aa^{-1} =1$
