For an integer $r$, let $A_r(x) = \#\{n\in \mathbb{N}: r\mid n \land 1\leq n\leq x\}$ be the function that returns the number of integers less than $x$ that are divisible by $r$. Then, the natural density of $S_r = \{n\in \mathbb{N}: r|n\}$ is defined as $$d(S_r) = \lim_{x\rightarrow \infty} \frac{A_r(x)}{x}$$
If $r$ is prime, we have $A_r(x) = \lfloor x/r \rfloor$, which implies that $\frac{x}{r}-1<A_r(x)< \frac{x}{r}$ and $d(S_r) = 1/r$
Using that same thought process, it seems like the natural density of $S_r$ for a random integer $r$ will be $1/r$. is that right? or am I missing something?
Considerations about "random" integers aside, you're not missing anything - your calculation holds for all $r$, not just when it's prime. The number of integers between $1$ and $N$ that are divisible by $r$ is always going to be $\lfloor \frac{N}{r} \rfloor$ and so you can bound $\frac{A_r(N)}{N}$ between two values that will both converge to $\frac{1}{r}$ in the limit.