Existence of a "quasi-uniform" probability distribution on $\mathbb{Z}$

No. Let's restrict our attention to $\mathbb{N}$. The hypotheses imply that if $q$ is a prime, then the probability that a random positive integer is not divisible by $q$ is $1 - \frac{1}{q}$. They also imply that these events are independent. Now let $n$ be a positive integer. If $q_1, q_2, \dots$ is an enumeration of the primes not dividing $n$ it follows that

$$p_n \le \prod_{i=1}^m \left( 1 - \frac{1}{q_i} \right)$$

for all $m$. But taking $m \to \infty$ the RHS approaches $0$; contradiction. Note that this argument does not require the prime number theorem; we just need to know that the harmonic series diverges.

Edit: Here is a generalization which more completely rescues Henry Cohn's argument. Generalize the condition to being that the probability of a positive integer being divisible by $n$ is $\frac{1}{n^s}$, for some real number $s > 0$. This is equivalent to requiring that the probability is 1) multiplicative in $n$ and 2) monotonically decreasing.

It follows that if $q$ is prime, then the probability that the exponent of $q$ in the prime factorization of a random positive integer is exactly $k$ is

$$\frac{1}{q^{ks}} - \frac{1}{q^{(k+1)s}} = \frac{1}{q^{ks}} \left( 1 - \frac{1}{q^s} \right).$$

We again have that for different primes $q$ these events are independent. Now, if $q_1, q_2, \dots$ is an enumeration of the primes and $n = \prod q_i^{k_i}$ is a positive integer, it follows that

$$p_n \le \prod_{i=1}^m \frac{1}{q_i^{k_i s}} \left( 1 - \frac{1}{q_i^s} \right).$$

for all $m$. If $s \le 1$ then the RHS converges to $0$ as $m \to \infty$ (this, again, does not require the prime number theorem) and we get a contradiction. If $s > 1$ then the RHS converges to

$$p_n = \frac{1}{n^s \zeta(s)}$$

and this is an equality because the RHS is the probability that a random positive integer has the same prime factorization as $n$.

There is a straightforward generalization where $\mathbb{N}$ is replaced by the set of nonzero ideals in the ring of integers $\mathcal{O}_K$ of a number field $K$ and the probability of an ideal being divisible by an ideal $I$ is $\frac{1}{N(I)^s}$, where we get the Dedekind zeta function instead.