It seems to me most likely to be true that every integer can be written in infinitely many ways as the difference between a pair of co-prime semiprimes and that the number of pairs with both members under $x$ is a simpley described fiunction. I say this by analogy with twin primes and their generalizations as described below. For primes it seems that proofs using the current methods are unlikely. For semi-primes there might be proofs but I would bet against it (but not very much.)

It is highly expected, but unknown, that there are infinitely many pairs of primes $p,q$ with $p-q=2.$ There has been some progress in recent years but no proof as of yet.

The number of primes up to $x$ is $\pi(x) \sim \frac{x}{\log x}$ The number of primes $p \lt x$ such that $p+2$ is also prime is conjectured to be $\sim \frac{Cx}{(\log x)^2}$ for a certain constant $C.$ There are simple heuristic arguments for this and the computational data is very good.

It has long been conjectured that every even integer $k$ is the difference of two primes in infinitely many ways. It is now known that there is at least one $k \lt 248$ for which it is true.

For $k=4$ the density for pairs $p,p+4$ should be almost exactly the same as for $k=2.$ However for $k=6$ it should be higher, after all, knowing that $p$ is prime and not tiny makes $p+6$ not only odd, but also a non-multiple of $3.$ The same constant that works for $2$ should be correct for any power of $2$ and the one which works for $6$ should also work for $k=2^a3^b.$ In general the constant for $k$ should depend only on the distinct primes dividing $k.$ Again, the computation evidence looks very strong.

The number of semiprimes up to $x$ is $\pi_2(x) \sim \frac{x \log \log x}{\log x}$ So perhaps the number of semi-prime pairs $s_1,s_2$ with $s_1-s_2=k$ is $\sim \frac{C x( \log \log x)^2}{(\log x)^2}$ where $C$ depends only on the prime divisors of $k$ or maybe for semiprimes that is less of a consideration. I've kind of ignored the relatively prime condition. Without it we could look at the odd prime divisors of $k$ (for $k$ even) pick any one odd divisor $p*$ then find (if possible) pairs $p,q$ with $p-q=\frac{k}{p*}$ and use semiprimes $pp*,qp*.$ Using relatively prime semi-primes would allow many more solutions.