The only such operations are those of the form $m\otimes n=mna$ for some fixed integer $a$. When $a=1$ this is ordinary multiplication, and when $a=0$ it’s trivial.
Suppose that $\otimes$ is such an operation. Then for any integer $n$ we have $$n\otimes 0=n\otimes(0+0)=(n\otimes 0)+(n\otimes 0)\,;$$ the only integer satisfying $x+x=x$ is $0$, so $n\otimes 0=0$ for every $n\in\mathbb{Z}$. Now let $a=1\otimes 1$. Then $$1\otimes 2=1\otimes(1+1)=(1\otimes 1)+(1\otimes 1)=a+a=2a\,,$$ and an easy induction shows that $1\otimes n=na$ for every positive integer $n$. We already know that $1\otimes 0=0$, so in fact $1\otimes n=na$ for every integer $n\ge 0$.
Now suppose that $n$ is a negative integer. Then $$0=1\otimes 0=1\otimes\big(n+(-n)\big)=(1\otimes n)+\big(1\otimes(-n)\big)=(1\otimes n)-na\,,\tag{1}$$ so $1\otimes n=na$, and we’ve now shown that $1\otimes n=na$ for every $n\in\mathbb{Z}$.
We can now repeat the argument to generalize the operand $1$ to any integer:
$$2\otimes n=(1+1)\otimes n=(1\otimes n)+(1\otimes n)=2an\,,$$ and another easy induction gives us $m\otimes n=mna$ for every non-negative integer $m$ and every integer $n$. Finally, the trick that I used in $(1)$ can be used again to show that $m\otimes n=mna$ for all $m,n\in\mathbb{Z}$.
Added: Because $\otimes$ is just scaled multiplication, it’s certainly both commutative and associative: $$m\otimes n=amn=anm=n\otimes m$$ and $$(k\otimes m)\otimes n=a(akm)n=ak(amn)=k\otimes(m\otimes n)$$ for any $k,m,n\in\mathbb{Z}$.
Best Answer
We can define $x \oplus y=y$. Then $(x \oplus y) \oplus z =z= x \oplus (y \oplus z)$ but $y=x \oplus y \neq y \oplus x=x$