Yes, for a finite set it's easy. For your example, $S=\{{2,5,17\}}$, $x$ has no factors in $S$ if and only if $x$ is relatively prime to $170=2\times5\times17$. The number of integers less than and relatively prime to 170 is $\phi(170)=1\times4\times16=64$, and then it's periodic beyond 170, so your limit is $64/170$, also known as $32/85$.
For infinite sets of primes, life is more difficult. I suspect that for the infinite set of primes in an arithmetic progression the limit is zero, that is, almost all big numbers have a factor in every arithmetic progression, but I can't provide a reference for this.
EDIT: A well-known example. The numbers that can be written as a sum of two squares, that's known to be a set of density zero (see, e.g., http://www.math.niu.edu/~rusin/known-math/93_back/2squares). Now those include all the numbers that have no prime factors of the form $4t+3$, so it follows that the set of numbers with no prime factor $4t+3$ has density zero.