Consider the pairing function $\pi:\mathbb{N\times N\to N}$ as given by wikipedia/Pairing_function, and let's just write $(k_1,k_2)\in\mathbb{N}\ $ for $\pi(k_1,k_2)$. Then the canonical definition, as given by wikipedia, doesn't sensibly interpret the sum $n=(i,j)+(k,l)$. That is, if we use the inverse construction to determine the "components" $(r,s)=n$ of the sum, then $i+k\neq r$, $j+l\neq s$, and there's no kind of sensible/intuitive algebraic relation whatsoever that I can see.
So is there any alternative definition of $\pi:\mathbb{N\times N\to N}$ such that, at least with respect to addition, "the sum of components equals the components of the sum", or some sensible algebraic relation between them?
There is not.
Note that $\pi$ being enumeration implies $(a, b) = (c, d)$ iff $a = c$ and $b = d$.
Let $x = (1, 2)$ and $y = (2, 1)$.
You require that $(na, nb) = n\cdot (a, b) = (a, b) \cdot n$, because $$n \cdot (a, b) =\\ (a, b) + (a, b) + \ldots + (a, b) = (a + a, b + b) + \ldots + (a, b) = (a + \ldots + a, b + \ldots + b) =\\(na, nb)$$
But then $(1 \cdot y, 2 \cdot y) = (1, 2) \cdot (2, 1) = (2 \cdot x, 1 \cdot x)$, then $y = 2x$ and $x = 2y$, which is impossible unless $x = y = 0$, but even then $\pi$ is not an enumeration.
More generally, you ask if $\mathbb N$ and $\mathbb N \times \mathbb N$ are isomorphic as semigroups. But semigroup rank (minimal number of elements of our semigroup such that minimal subsemigroup containing them is the semigroup itself) is preserved under isomorphism, while rank of $\mathbb N$ is $1$ and rank of $\mathbb N \times \mathbb N$ is $2$.