"Multi-distributive" operations on sets.

Question

As you can remember from Algebra courses, if $A$ is a set, and exist functions $\oplus , \odot: A \to A$ such that $(A,\oplus,\odot)$ is a ring, then such ring has a distributive property such that $$\forall \, a,b,c \in A \quad \; \, a \, \odot \, (b\, \oplus c) = (a \odot b) \; \oplus \; (a \oplus c ) $$ Or commonly said "$\odot$ distributes over $\oplus$".
My question is, ¿Is there any ring with distinct operations such that "$\oplus$ distributes over $\odot$"? Or more specifically: $$\forall \, a,b,c \in A \quad \; \, a \, \oplus \, (b\, \odot c) = (a \oplus b) \; \odot \; (a \oplus c ) $$ Notice that what I'm looking for is a ring with a pair of operations such that "$\odot$ distributes over $\oplus$" and "$\oplus$ distributes over $\odot$" simultaneously.
It's clear that at least in rings like $\mathbb{Z}$ or $\mathbb{Q}$ or even $\mathbb{Z_2}$ this doesn't happen with usual product and sum.
I'm wondering if this can even exist, a kind of "bi-distributivity" (clearly, if this comes out as true, my dream question could be: if you have $n$ distinct operations in a set, could you have a ${n \choose 2}$-distributivity for an operation ? .
Thanks for your time.

Answer 1

let $(R,+,\cdot,0,1) $ be a ring in which $'+'$ distributes $'\cdot '$ we show that $R=0$. (In the comment section Arturo shows this is true for rings without a multiplicative unit)

1. for any 3 elements $a,b,c\in R$ we have: $$a+bc=(a+b)(a+c)=a^2+ac+ba+bc$$ $$\implies (1):a=a^2+ac+ba$$
2. if a=b=c in (1) we get: $$ 3a^2=a$$
3. therefore if a=c in (1) we get: $$3a^2=a=2a^2+ba\implies a^2=ba$$
4. in (3) put a=1 and get: $$\forall b\in R:1=b$$