Is $\mathbb{Z}$ the initial rook?

Question

By a rook, let us mean a unital, not-necessarily associative near-ring satisfying $x0=0$.

Question. Is $\mathbb{Z}$ the initial object in the category of rooks?

(I hope so, since this is my only motivation for defining the concept of a rook.)

Here's an explicit list of axioms:

1. $x+0=x$
2. $0+x=x$
3. $(x+y)+z=x+(y+z)$
4. $-x+x = 0$
5. $x+-x=0$
6. $x1=x$
7. $1x=x$
8. $(x+y)z=xz+yz$
9. $x0=0$

Answer 1

If $R$ is a "rook", there is a unique homomorphism of groups $\mathbb{Z} \to R$ mapping $1$ to $1$, and hence $n$ to $n 1:=\underbrace{1+1+\dotsc}_{n}$. We only have to show that it is multiplicative, i.e. that $(n 1) (m 1) = (nm) 1$. Using 8. one reduces to the case that $n=1$, but then it follows from 7.

Since you want the axioms to be minimal: It is well-known that 5. follows from 1-4. For the proof that $\mathbb{Z}$ is initial we don't need 6. And we don't need 9.

PS: It doesn't make much sense to define a category of algebraic structures by random axioms and look for an initial object, when you don't know interesting examples of these structures yet. An example of a non-associative and non-distributive "rook" is the free "rook" on one generator $x$. Remark that here we cannot simplify $x(2x)$ or $x(x^2)$, which is quite nasty. I doubt that "rooks" are interesting ...