Show a property of a vector space axiom equivalent to another property Will try to prove additive inverse axiom is equivalent to saying $0v = 0$ for all $v \in V.$
Given $v+ w = 0$ for all $v$ and some $w \in V$, we have $w = -v$ and $v+ w = -v + v = -1v + 1v = (-1 + 1)v = 0v = 0$. 
In the other direction, $0 =  v – v = 0v$. Let $w = -v$. Then $0 = 0v = w + v$.
Checking if it works.
 A: $v+ (-v) = 0 \iff v\cdot 1 + v \cdot (-1) =0 ^ * \iff v(1+(-1)) = 0 \iff v\cdot 0 = 0$
$^*$Prove this separately. 
A: When showing that $0v=0$, you've assumed $-1v=-v$. This isn't an axiom and it needs to be proven before using this (correct) fact.
Here's a proof using just the axioms:
Let $V$ be a vector space and $v \in V$. We have $0v+0v=(0+0)v=0v$ (using the axiom that $(s+t)v=sv+tv$ for vectors $v$ and scalars $s,t$). But $0v$ is an element of $V$, so it must have an (additive) inverse: $-(0v)$. Add this to our previous equation: $(0v+0v)+(-0v)=0v+(-0v)$. Using associativity, $0v+(0v+(-(0v)))=0$ and so $0v+0=0$ and so $0v=0$ (here we've used the axiom which says $0$ is an additive identity several times). 
Now for the other direction, your proof again has a problem of assuming too much. You've assumed that $-v$ exists while showing that it exists!
Instead, assume that $0v=0$. Then $v+(-1v) = 1v+(-1v)=(1-1)v=0v=0$. Here I've used the axioms: $1v=v$, $(r+s)v=rv+sv$ and our assumption that $0v=0$. 
Likewise, $(-1v)+v=(-1v)+1v=(-1+1)v=0v=0$. Therefore, $w+v=v+w=0$ when $w=-1v$. This shows that $v$ has an additive inverse, namely $-1v$. If we also showed that when an additive inverse exists it is unique (i.e. there can be only one), then we would have $-1v=-v$. :)
