Commutative ring with unit is defined as $(R,+,\times)$, where $(R,+)$ is abelian group and $(R,\times)$ is commutative multiplicative monoid with $1$ and $+$ and $\times$ satisfies distributive law.
Could you give me an example $(R,+,\times)$ cannnot be a ring because $+$ and $\times$ does not satisfy distributive law although $(R，+)$ is abelian group and $(R,\times)$ is commutative multiplicative monoid with $1$.
Here is a "dumb" example. Let $R=\mathbb Z$, and let $\times=+$, i.e., addition and multiplication are the same thing. Now $(R,\times)$ is a commutative monoid, with a $1$ (i.e, $0\in R$). This is clearly not distributive: $1\times(1+1)=3\neq1\times 1+1\times 1=4$.
Let $(R,+) = \mathbb{Z}/3\mathbb{Z}$, and define $\cdot$ to be the unique commutative operation $R \times R \to R$ such that
$[0] \cdot a = [0]$ for all $a$
$[1] \cdot a = a$ for all $a$
$[2] \cdot [2] = [0]$
You can check that $(R,\cdot)$ is a commutative monoid, but the distributivity law does not hold because
$$[2] \cdot [1] + [2] \cdot [1] = [2] + [2] = [1] \neq [0] = [2] \cdot [2] = [2] \cdot ([1] + [1]).$$
+1, this is a great question. I was thinking about this a while back and asked this question which you might be interested in; https://math.stackexchange.com/questions/3900991/understanding-algebras-with-alteratives-to-the-distributive-law. Your question really nails what I was trying to think about there – Jojo Mar 21 '22 at 13:07