Does the Prime Number Theorem have anything to do with Erdos-Kac law or vice versa?

Yes. The number of prime factors of a number is distributed roughly like a Poisson process of expectation $\log \log n$, so the probability of exactly one prime factor is roughly $e^{- \log \log n} = 1/\log n$.

Remember that natural numbers of size roughly $n$ correspond, in the number field / function field dictionary, to polynomials of degree roughly $\log n$, whose prime factorizations are controlled by the permutation of Frobenius acting on the roots, a random permutation of $\log n$ letters, where primes correspond to cycles. You can check that the distribution of the number of cycles of a random permutation is close to a Poisson process because, e.g., writing permutations in standard cycle form, the chance of stopping a cycle after a given number of letters is determined only by number of previous letters and not the previous cycles, so the number of cycles is a sum of many independent random variables.

However this will not rigorously establish facts about numbers.

For your second question, I think you will find it very hard to find a natural set of numbers where the average number of prime factors changes, without doing something very drastic to the distribution. In general, one finds that the distribution of the number of prime factors is very insensitive to almost all conditions you can place on a number (e.g. congruence conditions), other than conditions stated directly in terms of the prime factorization.

Of course for a condition stated in terms of the prime factorization, there is no reason to expect that the number of prime factors and the probability of being prime should move in tandem.

The right context of your question is probably the area of Beurling primes. An arithmetic semigroup is a semigroup $(G,\cdot)$ together with a norm $\|\cdot\|:G\rightarrow[1,\infty)$, such that $\|gh\|=\|g\|\|h\|$ and for all $x$ we have that $N(x) = \#\{g\in G: \|g\|\leq x\}$ is finite. Define $\pi(x)$ as the number of indecompsable elements of $G$ with norm $\leq x$. Beurling proved that $N(x)=cx+\mathcal{O}(\frac{x}{x^{3/2+\delta}})$ implies the prime number theorem in the form $\pi(x)\sim\frac{x}{\log x}$, and the condition is best possible in the sense that for every $\delta>0$ there are groups satisfying $N(x)=cx+\mathcal{O}(\frac{x}{x^{3/2-\delta}})$ and $\pi(x)\not\sim\frac{x}{\log x}$.

The theory of Beurling primes has two motivations: On one hand it has applications to algebra and logic, see the books "Abstract analytic number theory" by Knopfmacher and "Number theoretic densities and logical limit laws" by Burris. On the other hand it serves as a context in which "equivalence" of statements about primes can be made precise.

In this context your question becomes: Is the set of arithmetic semigroups satisfying the prime number theorem equal to the set of semigroups satisfying Erdos-Kac? Erdos-Kac is essentially equivalent to $\sum_{p\leq x}\frac{1}{p}\sim\log\log x$, which is obviously implied by the prime number theorem, but Pollack ( http://pollack.uga.edu/beurling.pdf Wayback Machine ) showed that $\sum_{p\leq x}\frac{1}{p}\sim\log\log x$ is equivalent to the statement $\zeta_G(s)\sim\frac{A}{s-1}$, where $\zeta_G(s)=\sum_{g\in G}\|g\|^{-s}$. In particular all semigroups with $N(x)\sim Ax$ satisfy Erdos-Kac, while the counterexamples found by Beurling show that Erdos-Kac is strictly weaker then the prime number theory.