Suppose merely that $P$ is the sequence of primes starting with the $k$th and that $A$ is a sequence of integers such that $A_m$ has at least $m$ prime factors. Then $A_m \ge 2^m.$ The primes grow much more slowly so, no matter how large $k$ is, $P_n \lt A_{n+1}$ will quickly be impossible.
The $j$th prime $p_j$ is roughly $j \ln{j}$ and it seems a sure thing that $p_j<j^2$ after $p_1=2$. (Though it is probably not something proved or else the next result would be pointless.)
A weak but absolute bound which will suffice is that $p_j<1.2^j$ for $j \ge 26.$ This follows from the facts that $p_{26}=101 \lt 1.2^{26} \approx 114$ and that there is always a prime between $x$ and $\frac{6}{5}x$ for $x \ge 26$.
So that is a proof for the question asked. It is not clear to me how this advances the goal of finding a sequence " such that between every two consecutive elements there exists exactly one prime numbe.r"
There are such sequences such as $A_n=p_{k+n}+1$. But, as you suspect, none with the number of prime divisors increasing.
It also seems unpromising to have any formula which does not reference the primes. There are many open questions such as this one form 1882 about $\pi(x)$, the number of primes less than $x$. So we shouldn't expect to be able to answer the corresponding questions for a sequence $A_n.$