We will show that the set $S = \{k \cdot 5^{k-1} : k \in \mathbb{Z}^+\}$ is an appropriate example.

We first prove that $\mathbb{P}$ is $S$-proper for $N = 4$. Let $p \neq 5$ be prime and $0 \le n < p$. Then by Chinese remainder theorem there exists a positive integer $k$ satisfying the conditions $k \equiv n \pmod{p}$ and $k \equiv 1 \pmod{p-1}$. Therefore, by Fermat's theorem we obtain that $k \cdot 5^{k-1} \equiv n \pmod{p}$. For $p = 5$; as $1, 10 \in S$, $N = 4$ works.

Now suppose that $\mathbb{Z}^+$ is $S$-proper. Then there exists a positive integer $N$ satisfying the given conditions. Therefore we can get any residue in $(\text{mod } 5^N)$ as sum of at most $N$ elements of $S$. Since $k \cdot 5^{k-1} \equiv 0 \pmod{5^N}$ for all $k > N$, we have that for each integer $0 \le n < 5^N$, there exist non-negative integers $x_1, x_2, \dots, x_N$ such that
$$x_1 \cdot 1 + x_2 \cdot 2 \cdot 5 + \dots + x_i \cdot i \cdot 5^{i-1} + \dots + x_N \cdot N \cdot 5^{N-1} \equiv n \pmod{5^N}$$
and $x_1 + x_2 + \dots + x_N \le N$. Observe that for each $n$, we get a different $N$-tuple. On the other hand, the number of $N$-tuples of non-negative integers satisfying $x_1 + x_2 + \dots + x_N \le N$ is $\binom{2N}{N} < 5^N$ and hence we get a contradiction.