Distributive Property - A Converse Theorem 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 $a$.

$a+0=a$

$a \ast 0 = 0$

What other conditions (in addition to the above) are sufficient to imply a distributive property?
Listen, the converse of the distributive property does work. It is common in algebra for you to use the converse of the distributive property ex.) 2x + 3x = 5x or x(2 + 3).
Listen, the converse of the distributive property does work. It is common in algebra for you to use the converse of the distributive property ex.) 2x + 3x = 5x or x(2 + 3).