the principal is distribution: $a(b+c) = ab + ac$.
So the proof goes like this:
$x + (-1)x = 1\cdot x + (-1)\cdot x$ (by existence and definition of multiplicative identity)
$=(1+(-1))\cdot x$ (by distribution)
$=0\cdot x$ (by definition of additive inverse)
$=x\cdot 0$ (commutivity of multiplication but I have no idea why he did this)
$= 0$ (This is not an axiom but a proposition can be proven that $0\cdot x = 0$. Have you proven that yet? Does Spivak use that as an axiom?)
Then by definition we have that for every $x$ there exists a unique $-(x)$ so that $x + (-x) = 0$.
If we ever have an $a$ so that $x + a = 0$ it must be that $a=-x$ as the multiplicative inverse is unique. As $x + (-1)x =0$ it must be $(-1)x = -x$.
Prop: $x\cdot 0 = 0$.
Pf: $x\cdot 0 + (-(x\cdot 0)) = 0$. (Every element $a$, including $x\cdot 0$, has an additive inverse, $-a$, so that $a + (-a) =0$.)
$x\cdot(0 + 0) + (-(x\cdot 0)) = 0$ ($0=0+0$ because $0$ is the additive identity and $a +0 = a$ for all $a$, including when $a$ is $0$.)
$(x\cdot 0 + x\cdot 0) + (-(x\cdot 0)) = 0$ (distributivity)
$x\cdot 0 + (x\cdot 0 + (-(x\cdot 0)) = 0$ (associativity)
$x\cdot 0 + 0 = 0$ (definition of additive identity)
$x\cdot 0 = 0 $ ($a + 0= a$ for all $a$ by definition of additive identity.)