Measure on the positive integers respecting independence of prime divisors
Such measure doesn't exist. The idea is to show that the only measure satisfying 1 (which, by the way, obviously implies 2) is $$ \mu(\{n\}) = \frac{1}{n\zeta(1)} = 0, n\ge 1, $$ which is a contradiction.
So let $\mu$ satisfy 1. Denote by $\mathbb{P}$ the set of primes. Then, for any $n\in \mathbb N$, we have by the inclusion-exclusion formula (all series below are divergent, but one can proceed by finite approximations) $$ \mu(\{n\}) = \mu(A_n) - \mu\Big(\bigcup_{p\in \mathbb{P}} A_{np}\Big) \\ = \frac{1}{n} - \sum_{p\in \mathbb P}\mu(A_{np}) + \sum_{\substack{p_1,p_2\in \mathbb P\\p_1\neq p_2 }}\mu(A_{np_1}\cap A_{np_2}) - \sum_{\text{distinct } p_1,p_2,p_3\in \mathbb P}\mu(A_{np_1}\cap A_{np_2}\cap A_{np_3}) + \dots \\ = \frac1n + \sum_{k=1}^\infty (-1)^k \sum_{\text{distinct } p_1,p_2,\dots,p_k\in \mathbb P}\mu\Big(\bigcap_{j=1}^k A_{np_j}\Big)\\ = \frac1n + \sum_{k=1}^\infty (-1)^k \sum_{\text{distinct } p_1,p_2,\dots,p_k\in \mathbb P}\mu\Big( A_{np_1p_2\cdots p_k}\Big) \\ = \frac1n + \sum_{k=1}^\infty (-1)^k \sum_{\text{distinct } p_1,p_2,\dots,p_k\in \mathbb P}\frac1{np_1p_2\cdots p_k} = \frac1n\prod_{p\in \mathbb P}\Big(1-\frac{1}{p}\Big) = 0, $$ as claimed.
Similarly, it can be shown that the unique measure satisfying $\mu_s(A_n) = n^{-s}$ with $s>1$ is $$ \mu_s(\{n\})=\frac1n\prod_{p\in \mathbb P}\Big(1-\frac{1}{p^s}\Big) = \frac{1}{n^s \zeta(s)}, n\ge 1. $$
Such a measure can not be a probability (i.e. we can't have $\mu(\mathbb{N})=1$).
Ad absurdum, suppose we have a probability $P$ satisfying $(1)$, $(2)$.
Let $(p_n)_{n \ge 1}$ be an ordered enumeration of prime numbers. There is no positive integer which has an infinity of prime divisors, and thus $$\bigcap \limits_{n \in \mathbb{N}} \bigcup \limits_{k \ge n} A_{p_k} = \varnothing.$$
Hence $P \Big(\bigcap \limits_{n \in \mathbb{N}} \bigcup \limits_{k \ge n} A_{p_k} \Big) = 0$, so $$P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) > 0.$$
However, $P$ is continuous below so \begin{align*} P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) & = \lim \limits_{n \to \infty} P \Big( \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) \\ & = \lim \limits_{n \to \infty} \prod \limits_{k \ge n} P (\overline{A_{p_k}}) \end{align*}
because $P$ is a probability, the $A_{p_k}$ are mutually independant, and thus so are the $\overline{A_{p_k}}$. Hence \begin{align*} P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) & = \lim \limits_{n \to \infty} \prod \limits_{k \ge n} \Big( 1 - P (A_{p_k}) \Big) \\ & \le \lim \limits_{n \to \infty} \prod \limits_{k \ge n} e^{-P(A_{p_k})} \end{align*}
because $1-x \le e^{-x}$ (convexity inequality). We deduce finally :
$$P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) \le \lim \limits_{n \to \infty} e^{-\sum \limits_{k=1}^n P(A_{p_k})}$$
But $$\sum \limits_{k=1}^n P(A_k) = \sum \limits_{k=1}^n \frac{1}{p_k} \underset{n \to \infty}{\longrightarrow} +\infty$$
(see Divergence sum of reciprocal of primes), so we conclude
$$P \Big( \bigcup \limits_{n \in \mathbb{N}} \bigcap \limits_{k \ge n} \overline{A_{p_k}} \Big) = 0$$
which is absurd. Hence, there is no probability satisfying your conditions.
See Zhoraster's post which provides a better and clearer answer to your question.