You can see below an argument that $A_n$ can't have a strictly increasing number of prime divisors. I like it but it was the result of not reading the question carefully. However I still wonder how your question advances your goal.

All I have to add is an observation and computation. The observation is that the (or at least a) main problem seems to be with twin primes. There is no proof that there are infinitely many, but no one doubts it. Then there is some fluctuation in the number of prime divisors of the one number between the two primes.

I somewhat arbitrarily looked] at the next $5000$ primes after $200,000,000.$ If $p_1,p_2,P_3$ are three consecutive primes and every integer in $[p_1+1,p_2-1$ has more prime divisors then anything in $[p_2+1,p_3-1]$ then there will be an immediate obstacle to starting the $A$ sequence before $p_2.$ I found $70$ cases where that happened. all of them were for $p1,p2$ a twin prime pair. There were $326$ such pairs.

A typical example is $200085887, [10], 200085889, [3, 3, 3, 3, 4, 3, 5, 4, 3, 3, 7, 2, 3, 6, 7, 3, 4, 4, 5, 4, 4, 2, 8, 3, 4,$$ 4, 3, 3, 5, 4, 8, 4, 3, 3, 5, 5, 3, 2, 5, 2, 7, 2, 3, 3, 3, 5, 8, 3, 3, 3, 4, 3, 5, 2, 4, 4, 3, 2, 6, 3, 4, 3, \mathbf{9}, 2, 5, 3, 5, 6, 2, 2, 8]$

This means that between the two primes indicated there is one integer and it has $10$ prime factors (with repetition) . Even though it is a pretty long way to the next prime, none of the integers in that range have more than $9$ prime factors.

**OLD ANSWER** 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 $j \ln{j}.$ Here is a reference for the claim $p_j \lt j(\ln{j}+\ln\ln{j}-9.4)$ for $j \ge 15985.$

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 number"*

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 $A_n.$