This is not a complete answer, but I want to share my progress.

Let $\mathcal{C} \subseteq \mathbf{Set}$ be the category of sets consisting of all $\mathbb{N}^k$ with $k \in \mathbb{N}$. It is cartesian. It follows that for every symmetric monoidal category $\mathcal{D}$ the category of symmetric monoidal functors $\mathcal{C} \to \mathcal{D}$ is equivalent to the category of (finite product)-preserving functors $ \mathcal{C} \to \mathrm{CocommCoAlg}(\mathcal{D})$. If $\mathcal{D}$ is cartesian (as in your example), then the forgetful functor $\mathrm{CocommAlg}(\mathcal{D}) \to \mathcal{D}$ is an isomorphism of categories.

Thus, it suffices to find the product-preserving functors $F : \mathcal{C} \to \mathcal{D}$. (You only asked for $\mathcal{D} = \mathbf{Set}$, but I don't think that this makes the problem easier. Let us keep $\mathcal{D}$ arbitrary for now.) This essentially means to find "all the structure" on $\mathbb{N}$ in the language of cartesian categories. Namely, any structure on $\mathbb{N}$ leads to a structure on $X := F(\mathbb{N}) \in \mathcal{D}$, and clearly the object-part of $F$ is determined by $X$.

But there are actually a lot of structures. First of all, $\mathbb{N}$ carries the structure of a commutative semiring. It follows that $X$ also carries the structure of a commutative semiring. Also, for every map $f : \mathbb{N} \to \mathbb{N}$ we must have a map $f_* : X \to X$, and surely $\mathrm{id}_* = \mathrm{id}$ and $(fg)_* = f_* g_*$. But this also must be true for all maps $f : \mathbb{N}^k \to \mathbb{N}^n$, which induce morphisms $X^{\otimes k} \to X^{\otimes n}$. For example, there is an isomorphism $\mathbb{N} \cong \mathbb{N} \times \mathbb{N}$, thus inducing an isomorphism $X \cong X \otimes X$.

I should stop here. I don't think that there is any reasonable description.

Here is another approach, but I am not sure if it is useful. $\mathcal{C}$ is equivalent to a category $\mathcal{C}'$ which has exactly two objects, $1$ and $\mathbb{N}$. Choose an isomorphism $\delta : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ (for example, $\delta(n,m) := 2^n (2m+1) - 1$). Then $\mathbb{N} \times \mathbb{N} = \mathbb{N}$ in $\mathcal{C}'$, the projections are $p_1 \circ \delta^{-1}$ and $p_2 \circ \delta^{-1}$. The category of (finite product)-preserving functors $\mathcal{C} \to \mathcal{D}$ is equivalent to the category of (finite product)-preserving functors $F : \mathcal{C}' \to \mathcal{D}$. On objects, they are determined by $X := F(\mathbb{N})$ again. And again we have the morphisms $f_* : X \to X$ for all maps $f : \mathbb{N} \to \mathbb{N}$. And we need that $$((p_1 \circ \delta^{-1})_*,(p_2 \circ \delta^{-1})_*) : X \to X \times X$$ is an isomorphism. Probably much more is needed.

This is not a complete answer, but I want to share my progress.

Let $\mathcal{C} \subseteq \mathbf{Set}$ be the category of sets consisting of all $\mathbb{N}^k$ with $k \in \mathbb{N}$. It is cartesian. It follows that for every symmetric monoidal category $\mathcal{D}$ the category of symmetric monoidal functors $\mathcal{C} \to \mathcal{D}$ is equivalent to the category of (finite product)-preserving functors $ \mathcal{C} \to \mathrm{CocommCoAlg}(\mathcal{D})$. If $\mathcal{D}$ is cartesian (as in your example), then the forgetful functor $\mathrm{CocommAlg}(\mathcal{D}) \to \mathcal{D}$ is an isomorphism of categories.

Thus, it suffices to find the (finite product)-preserving functors $F : \mathcal{C} \to \mathcal{D}$. (You only asked for $\mathcal{D} = \mathbf{Set}$, but I don't think that this makes the problem easier. Let us keep $\mathcal{D}$ arbitrary for now.) This essentially means to find "all the structure" on $\mathbb{N}$ in the language of cartesian categories. Namely, any structure on $\mathbb{N}$ leads to a structure on $X := F(\mathbb{N}) \in \mathcal{D}$, and clearly the object-part of $F$ is determined by $X$.

But there are actually a lot of structures. First of all, $\mathbb{N}$ carries the structure of a commutative semiring. It follows that $X$ also carries the structure of a commutative semiring. Also, for every map $f : \mathbb{N} \to \mathbb{N}$ we must have a morphism $f_* : X \to X$, and surely $\mathrm{id}_* = \mathrm{id}$ and $(fg)_* = f_* g_*$. But this also must be true for all maps $f : \mathbb{N}^k \to \mathbb{N}$, which induce morphisms $X^{k} \to X$. For example, there is an isomorphism $\mathbb{N} \cong \mathbb{N} \times \mathbb{N}$, thus inducing an isomorphism $X \cong X \times X$.

I should stop here. I don't think that there is any reasonable description.

Here is another approach, but I am not sure if it is useful. $\mathcal{C}$ is equivalent to a category $\mathcal{C}'$ which has exactly two objects, $1$ and $\mathbb{N}$. Choose an isomorphism $\delta : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ (for example, $\delta(n,m) := 2^n (2m+1) - 1$). Then $\mathbb{N} \times \mathbb{N} = \mathbb{N}$ in $\mathcal{C}'$, the projections are $p_1 \circ \delta^{-1}$ and $p_2 \circ \delta^{-1}$. The category of (finite product)-preserving functors $\mathcal{C} \to \mathcal{D}$ is equivalent to the category of (finite product)-preserving functors $F : \mathcal{C}' \to \mathcal{D}$. On objects, they are determined by $X := F(\mathbb{N})$ again. And again we have the morphisms $f_* : X \to X$ for all maps $f : \mathbb{N} \to \mathbb{N}$. And we need that $$((p_1 \circ \delta^{-1})_*,(p_2 \circ \delta^{-1})_*) : X \to X \times X$$ is an isomorphism. Probably much more is needed.