Extending $\Bbb N$ to a semiring with isomorphic additive and multiplicative structure 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 $R$. Here, $\Bbb N$ is meant to include $0$. I do not think that there is such an extension, but I cannot find a contradiction. I also wonder if the problem becomes easiert when asking for an isomorphism 
$$\varphi:(R\setminus\{0\},\cdot)\cong(R,+)$$
instead?

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).
 A: There is an extension $R$: take the closure of $\mathbb N$ by the operations $\text{L}$ (or $\varphi$ in the OP) and its inverse $\text{E}$, which are the logarithm and exponential in base $1.2$. Notice that the choice of base (see below) implies that all new numbers generated by repeated applications of the 4 operations ($+$, $\cdot$, $\text{L}$ and $\text{E}$) are $\gt 1$, except for infinities, so that there is never a need to apply $\text{L}$ to negative numbers.
For infinities impose the following rules:
$0\cdot x=0$
$\text{L}(0)+x=\text{L}(0)$
$\text{L}(0)\cdot x=\text{L}(0)$ if $x\ne0$
$\text{L}(\text{L}(0))+x=\text{L}(\text{L}(0))$ if $x\ne \text{L}(0)$
$\text{L}(\text{L}(0))\cdot x=\text{L}(\text{L}(0))$ if $x\ne \text{L}(0),0$
$\text{L}(\text{L}(\text{L}(0)))+x=\text{L}(\text{L}(\text{L}(0)))$ if $x\ne \text{L}(\text{L}(0)),\text{L}(0)$
...
These rules obey the semiring axioms and cover all possibilities for mixing reals and infinities. Notice also that $\text{L}(0)$, $\text{L}(\text{L}(0))$, $\text{L}(\text{L}(\text{L}(0)))$ etc. are then the only non-reals in $R$.
As for the choice of a base $\lt\sqrt{2}$, notice that if $a=1.2577...$ is the solution to $1.2^a=a$, then both $\text{L}(x)>a$ and $\text{E}(x)>a$ if $x>a$ and therefore the only finite elements in this model $R$ that can be $<a$ are $0$, $1$, $\text{E}(1)$, $\text{E}(\text{E}(1))$, $\text{E}(\text{E}(\text{E}(1)))$ etc.
Update. A much simpler model is the completion of the one above, but with $\sqrt{2}$ as the base:
$ \text{L(x)}=\log_{\sqrt{2}}(x)$, $\text{E}(x)=\left(\sqrt{2}\right)^{x}$ its inverse, and
$R= \{\dots\ \text{L}^3(0), \text{L}^2(0), \text{L}(0), 0, 1, \sqrt{2}, \sqrt{2}^\sqrt{2},\sqrt{2}^{\sqrt{2}^\sqrt{2}}\dots\}\cup\{x\in\mathbb R| x\ge 2\}$
with the same rules for infinities as already explained.
