Multiplicative inverse of $0$

$$0\cdot a= (0+0)\cdot a=0\cdot a+0\cdot a$$ Now substract $(0\cdot a)$ on both sides (we can do this because $\forall r\ \exists(-r)\mid r-r=0)$:$$0=0\cdot a$$ This means that $0$ can only have a multiplicative inverse if $0\cdot \tilde a=0=1$ for some $\tilde a$. This then implies that we have $x=x\cdot1=x\cdot0=0\ \ \forall x$, hence we live in the zero-ring $\{0\}$.

So for any ring $R$, we have that $0$ has a multiplicative inverse $\iff$$R=\{0\}$

Suppose there exists $a = 0^{-1}$. Then $$ 1 = 0\cdot a = (a-a)\cdot a = a^2 - a^2 = 0 $$

Then, for all $b$ in the ring, $$b = b\cdot 1 = b \cdot 0 = 0 $$ so zero is the only element of the ring.

Consequence: If there is nonzero element of the ring, $0$ does not have a multiplicative inverse.

Distributivity implies absorption into the additive identity, i.e. $$\forall a,b,c \in R\quad a(b+c)=ab+ac \implies ab=a(b+0)=ab+a0 \implies a0=0.$$ So if everything times zero equals zero, then obviously you can't have something times zero equal one, unless of course zero equals one and you are looking at the zero ring.