Suppose that $\ast$ distributes over $+$, where $0$ is the additive identity. We can conclude the following.
$a \ast b = a \ast \left( b + 0\right)$
$a \ast b = a \ast b + a \ast 0$
$\therefore a \ast 0 = 0$
In other words, the additive identity and multiplicative 'annihilator' become equivalent under the distributive property. However, the converse is not true. Equating the two $0$'s is a necessary but insufficient condition to establish distributivity.
My question: suppose we know that the following two statements are true for any element $a$.
$a \ast 0 = 0$
What other conditions (in addition to the above) are sufficient to imply a distributive property?