What is an instructive example of a set $X$ equipped with two monoid structures $(X,+,0)$, $(X,\cdot,1)$, such that $+$ is commutative, the distributive laws hold, but $0 \cdot x = 0$ or $x \cdot 0 = 0$ do not hold?
Notice that in case these two absorbing laws hold, one calls $(X,+,0,\cdot,1)$ a semiring. At first sight, it might be surprising that these laws have to be imposed, but this is quite natural from a more general point of view, namely the multiplication $\cdot : X \times X \to X$ should be an additive monoid homomorphism in each variable, and it is known that one has to demand that monoid homomorphisms preserve the neutral element.
Let $S$ be the set of pairs in $\Bbb R^2$ of the form $[a,b]$ where $a\leq 0$ and $b\geq 0$.
Define $$[a,b]+[a',b']=[\min(a,a'),\max(b,b')]$$ and
$$[a,b][a',b']=[a+a',b+b']$$
All it takes for distributivity are the identities $\min(a,a')+c=\min(a+c,a'+c)$ and $\max(b,b')+c=\max(b+c,b'+c)$.
It turns out that $e=[0,0]$ is the neutral element for both operations of $S$, so $a=a+e=ae$, and then it's impossible for $e$ to multiplicatively absorb non $e$ elements.