Homomorphism of a set to its power set.

I'll assume you meant $\mathbb{N}$ everywhere. Then the set of such homomorphisms is pretty limited. For every $n \geq 1$ we have $f(n) = f(1)$:

$$f(n) = f(1 + \ldots + 1) = f(1) \cup \ldots \cup f(1) = f(1)$$

Since by definition $f(0) = \emptyset$ we are only left with one degree of freedom by setting $f(1)$ -- depending on whether you see restricting $f(1)$ included in your homomorphism definition.

The above equality can also be seen in a different, more general light. Every semiring homomorphism $f$ gives rise to a monoid homomorphism $$f_\mathrm{Mon}: (S, +) \to (S', \oplus)$$ wrt. the additive operation. This is just because a semiring "contains" a monoid. The additive monoid here is $(\mathbb{N}, +)$, which is generated by $1\in\mathbb{N}$. This in turn means that $f_\mathrm{Mon}$ is already fully determined by its image of $0$ and $1$. Since we have $f = f_\mathrm{Mon}$, we can conclude the same for $f$.
In fact you could have made similar considerations for the multiplicative monoid to conclude that $f$ is determined by its image on $0$, $1$ and primes.

Let $f: (\Bbb Z, +, \cdot, 0 ) \to(P(\Bbb Z), \cup, \cap, \emptyset))$ be a semiring morphism. Then $f(0) = \emptyset$ and for all $n \in \Bbb N$, $f(n + (-n)) = f(0) = \emptyset = f(n) \cup f(-n)$. It follows that $f(n) = f(-n) = \emptyset$. Therefore, $f(x) = \emptyset$ for all $x \in \Bbb Z$.

If you want that $f$ preserves the identity of the multiplication, you get $f(1) = \Bbb Z$, which contradicts the previous result.