Seen $(\Bbb N,+,\cdot)$ as a semiring, is it possible to extend it to a semiring $(R,+,\cdot)$ so that the additive and multiplicative monoids become isomorphic? This means there is some monoid-isomorphism

$$\varphi:(R,\cdot)\cong(R,+)$$

and $\Bbb N$ is a sub-semiring of $R$. 

If I had written $(R\setminus\{0\},\cdot)$ instead, then a solution would be $R=\Bbb R^{\ge 0}$ and $\varphi=\log$. But I want to include the zero. I do not think that there is such an extension, but I cannot find a contradiction.

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**Observations**

The multiplication will be commutative. Also, there will be *many* new "numbers", e.g. a unique additively absorbing element $\eta:=\varphi(0)$, i.e. $\eta+x=\eta$ for all $x\in R$. For $n\in\Bbb N^+$ we have

$$n\cdot \eta=\underbrace{\eta+\cdots+\eta}_n=\eta.$$

Therefore we have further elements $\tilde\eta=\varphi(\eta)$ that absorbe *some* numbers when added to them, e.g. $\varphi(n),n\in\Bbb N^+$, but not all (there can be only one universally absorbing element).