Ring without distributive law?

Thanks to Tegiri Nenashi for his comment suggesting using operations other than plus. This answer will use max. If we wish to have our $\boxplus$ be the multiplicative operation on a semiring then the additive identity needs to annihilate the semiring multiplicatively, thus the additive identity must be $0$. So the additive operation, $\max$, must be applied on nonnegative reals only. If we want the multiplicative identity, $\infty$, to be in the semiring also then we must take the underlying set to be $[0,\infty]$.

Inspired by reasonable intuitions and limiting arguments define $\max(a,\infty)=\infty$ and $a\boxplus \infty =a$ for all $a \in [0,\infty]$.

Then I claim that $([0,\infty],\max,0,\boxplus,\infty)$ is a commutative semiring. It is easy to check that $([0,\infty],\max,0)$ is a commutative monoid with identity $0$. Properties 1, 4 and 5 show that $([0,\infty],\boxplus,\infty)$ is a commutative monoid with identity $\infty$. Property 2 shows that $0$ annihilates $[0,\infty]$.

Finally we need the distributive law $$a \boxplus \max(b,c)=\max(a\boxplus b,a\boxplus c).$$

Without loss of generality $b \leq c$ so we need to show that $a \boxplus c=\max(a\boxplus b,a\boxplus c)$ i.e. $a\boxplus b \leq a\boxplus c$. If $a=0$ or $a=\infty$ or $b=c$ this is true, otherwise we need $$\frac{1+e^{a+b}}{e^a+e^b}\leq \frac{1+e^{a+c}}{e^a+e^c}$$ i.e. $$e^a+e^c+e^{2a+b}+e^{a+b+c} \leq e^a+e^b+e^{2a+c}+e^{a+b+c}$$ i.e. $$e^{2a}(e^b-e^c) \leq e^b-e^c.$$ But $e^b-e^c < 0$ so our inequality is equivalent to $e^{2a} \geq 1$ which is true.