I think you're step 2 is incorrect, if you're working with ring axioms, you can say in general for a number $\displaystyle a\in R$ (where R is a ring), there exists a neutral element $\displaystyle e$ (and for the addition: 0) wherefore: $\displaystyle a+e=a=e+a$, that means in this case (for the addition): $\displaystyle -0+0=-0$.

I would prove it this way:

$\displaystyle 0+0=0$ (this is an axiom of the existence of a neutral element in a commutative group in a ring)

$\displaystyle \Leftrightarrow 0=0-0$ (putting 0 to the other side)

$\displaystyle \Leftrightarrow 0=-0+0 $ (in a ring the addition is commutative)

Because of the existence of the neutral element $\displaystyle 0$ for the addition, like I stated earlier: $\displaystyle -0+0=-0$ therefore:

$\displaystyle 0=-0+0 \Leftrightarrow 0=-0$