Say we have an algebra $(A, +, \cdot)$, where $(A, +)$ is an Abelian Group. All we know about $\cdot$ is that it is both left and right distributive over addition. So, $\forall a,b,c \in A, a \cdot (b+c) = (a \cdot b) + (a \cdot c)$ and $(b+c) \cdot a = (b \cdot a) + (c \cdot a)$.
We can't assume $ \cdot $ is associative, commutative, or anything else besides distributive.
Do we know whether multiplication is power associative or not? That is, for all $a$, powers of $a$ are associative (e.g $a \cdot (a \cdot a) = (a \cdot a) \cdot a$).
If so, what would the proof look like? If not, is there a counterexample?
I attempted this myself, but I couldn't find any hints of a proof, so I tried to produce a counterexample, similarly with no luck.
It seems to me you can construct an example by doing something like a group ring but not with a group, instead with a non-power-associative magma.
Let $M$ be any magma that isn't power-associative. You can just use the free magma on one element, whose elements start out $\{x, xx, x(xx), (xx)x, \ldots\}$.
Then, form formal combinations using integers as coefficients $\sum_{m\in M} z_mm$ which are stipulated to be finitely supported.
Multiplication is defined distributively, so the resulting algebra should be distributive, but it is also clearly not power-associative since $(xx)x\neq x(xx)$.