Semigroup isomorphism between $(\{1,2,\dots \},\times)$ and $(\{0,1,2,\dots \},+)$. I know that the two semigroups $(\{0,1,2,\dots \},\times)$ and $(\{0,1,2,\dots \},+)$ are not isomorphic because if we want to map identity elements together then it can be see that  we can't have injective function between them,but what can we say about $(\{1,2,\dots \},\times)$ and $(\{0,1,2,\dots \},+)$?

As you have noted, it is easy to see that $(\mathbb{N}, +)$ and $(\mathbb{N}, \cdot)$ are not isomorphic as semigroups, since the latter possesses an algebraic feature that the first one does not, namely an absorptive element (a $0$-element, which in our case is precisely the natural number $0$). 
When comparing however $(\mathbb{N}, +)$ with $(\mathbb{N}^{*}, \cdot)$, it is important to realize that not only the former but also the latter are free commutative monoids, the former of basis $\{1\}$ and the latter of basis $\mathbb{P}$, the set of all prime numbers. Free commutative monoids have the following remarkable property of rigidity: 


*

*if $X$ is an arbitrary set and $\iota: X \to \mathbb{N}^{(X)}$ is the canonical injection into the free commutative monoid over $X$, then any generating system $S$ of $\mathbb{N}^{(X)}$ as a monoid will necessarily include the canonical basis $\iota(X)$.

*as a consequence, $\mathbb{N}^{(X)}$ has no other basis than $\iota(X)$.

*more generally, any free commutative monoid will have a unique basis and will thus admit a notion of dimension, i.e. the cardinality of the unique basis it possesses.


The dimension of a free commutative monoid is obviously an invariant under monoid isomorphisms; since $\mathrm{dim}\ \mathbb{N}^{*}=|\mathbb{P}|=\aleph_0$ and $\mathrm{dim}\ \mathbb{N}=1$ we now have a clear understanding of the obstruction to the existence of any isomorphism between the two monoids at hand.
