Rig in which 0 is not an absorber.

Let $(R,.,1)$ be a commutative monoïd such that $r^2=r$ for all $r\in R$ (for example, you can start with any Boolean ring $(R,+,.,0,1)$. Concrete example are given, by $(\mathbb{Z}/2\mathbb{Z})^{I}$ for any nonempty set $I$, for example )

Then set $+=.$ and $0=1$. Hence $(R,+,0)=(R,.,1)$ is a commutative monoïd. Moreover, for all $r,s,t\in R$, we have $r.(s+t)=r(st)=rst$ and $r.s+r.t=(rs)(rt)=r^2st=rst$, so you have the distributivity property. Now $r.0=r.1=r$ for all $r\in R$.

Let $E$ be the set of intervals of the real line of the form $[a,b]$ where $a\leq 0$ and $b\geq 0$.

Define $[a,b]\oplus [c,d]=[\mathrm{Min}(a,c), \mathrm{Max}(b,d)]$ and $[a,b]\otimes [c,d]=[a+c, b+d]$.

One can check that $(E, \oplus)$ and $(E,\otimes)$ are abelian monoids, both having identity $[0,0]$, and furthermore $\otimes$ distributes over $\oplus$.

If it had an absorbing zero element, it would be a full-fledged semiring... but it obviously does not, since the $\oplus$ identity is the same as the $\otimes$ identity, it is obviously not absorbing.

I found this example a long time ago in

Minoux, Michel, and Michel Gondran. Graphs, Dioids and Semirings. New Models and Algorithms. Vol. 41. Springer, 2008.

as example 5.3.1.