You missed the excitement of 1998, with $1999=1999, 2000=2^4 \cdot 5^3, 2001=3 \cdot 23 \cdot 29,$ and $2002=2 \cdot 7 \cdot 11 \cdot 13$
A quick set of thoughts, too long to fit in a comment:
this requires finding a "prime gap" of length $k-1$, since $s(n+1)=1$ means that $n+1$ is prime that $n+1$ is either prime or a power of a prime, but the next $k-1$ digits are composite since s(n+x)>1 for $2 \le x \le k$. This also means that $s(n+2)=2$ only because $s(n+2)$ is even, thus $2$ is one of the factors and implies that $(n+2)/2$ is prime (or that $(n+2)/2^j$ is prime for some $j \in \mathbb{Z}$), since $n+2$ only has two factors and one of them is $2$ (or $2^j$).
For $s(n+k)$ to have $k$ distinct prime factors means that it has to be at minimum a product of the first $k$ prime numbers, while it definitely has to be a multiple of the product of $k$ prime numbers. So the two key restrictions are that s(n+k) is $k$-composite (has $k$ prime factors) and that both (n+1) and (n+2)/2 are prime numbers.
Hmm, I thought something about the fact that either s(n+2) or s(n+4) would be divisible by $4$ while the other would be divisible by $2$ but not by $4$ would play some role in this.
Here are some quick results from running "bash", "factor", and "sed" and "awk" at the command line:
If you want an ascending run of 1,2, and 3 prime factors, the smallest example starts at $n=63$, with $64=2^6, 65=5 \cdot 13, 66=2 \cdot 3 \cdot 11$
If you want an ascending run of 1,2,3, and 4 prime factors, we already missed the exciting years of $n=1866$ and $n=1998$
1867: 1867
1868: 2 2 467
1869: 3 7 89
1870: 2 5 11 17
1999: 1999
2000: 2 2 2 2 5 5 5
2001: 3 23 29
2002: 2 7 11 13
And the next few years with ascending runs of 1,2,3, and 4 prime factors will start after the years 3216, 4056, and 4176 with 3217, 4057, and 4177 as prime years. Unfortunately, these computational results are not giving me the germ of any shortcut or understanding. There are also some descending sequences in terms of the number of prime factors, and their placement also does not help.
If you want an ascending run of 1,2,3,4, and 5 prime factors, we have to wait almost half-a-million years to get to the exciting years of $n=491850$ and $n=521880$ for $k=5$
491851: 491851
491852: 2 2 122963
491853: 3 19 8629
491854: 2 11 79 283
491855: 5 7 13 23 47
521881: 521881
521882: 2 260941
521883: 3 3 3 3 17 379
521884: 2 2 11 29 409
521885: 5 7 13 31 37
Now with four numbers computed and found, I searched the OEIS and found the corresponding sequence. Since the Online Encyclopedia already has this sequence, I'm hanging up my computational hat and heading off to work. :)
http://oeis.org/A086560
Start of first run of n successive numbers in which i-th number has exactly i distinct prime divisors for i = 1..n
2, 5, 64, 1867, 491851, 17681491, 35565206671
J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 64, p. 23, Ellipses, Paris 2008.
Heuristically this should be the case. For any prime p greater than 5, consider the set of numbers of the form $2^a3^b5^c p \pm 1$. The "probability" that one of of these is prime should be about $$\frac{1}{\log (2^a 3^b 5^c)} = \frac{1}{a \log 2 + b \log 3 + c \log 5} .$$ So the probability that both $2^a3^b5^c p + 1$ and $2^a3^b5^c p - 1$ are prime should be about $$\frac{1}{\left(a \log 2 + b \log 3 + c \log 5\right)^2}.$$ Now, note that the product $$\prod_{a,b,c} \left(1-\frac{1}{\left(a \log 2 + b \log 3 + c \log 5\right)^2}\right)$$ diverges to $0$, it represents the probability that none of those $2^a3^b5^c p \pm 1$ is a prime pair. So we should expect for a given $p$ there should be such an $a$,$b$ and $c$. So after noting the twin prime pairs $(3,5)$, $(5,7)$ and $(29,31)$ it seems like we should expect a much stronger statement. For any prime $p$, there should be a positive integer $n$ such that $p$ is the largest prime divisor of $n$ and $n+1$ and $n-1$ are both prime.
Obviously, proving something like this is well beyond current technology. I'd also say that it is highly likely that even your weaker statement is well beyond what is currently doable.
Best Answer
Assuming the prime tuples conjecture, all of these questions have affirmative answers. For instance, one can use the Chinese remainder theorem to find $a,b$ such that the tuple $$ an+b, \frac{an+b+1}{2^2 \times 5}, \frac{an+b+2}{3 \times 7}, \frac{an+b+3}{2 \times 11}, an+b+4$$ (for instance) are an admissible tuple of linear forms (in that they have integer coefficients, and for each prime $p$ there exist a choice of $n$ in which all forms are simultaneously coprime to $p$); concretely, one can take $a=2^2\times 3 \times 5 \times 7 \times 11 = 4620$ and $b=19$. The tuples conjecture then shows that there are infinitely many $n$ in which these forms are all simultaneously prime, giving infinitely many gaps of order $3$ and length $4$. Similarly for other gap orders and lengths.
On the other hand, unconditionally not much can be said. I think at our current level of understanding we cannot even rule out the ridiculous scenario in which every sufficiently large prime gap $(p_n,p_{n+1})$ contains a semiprime. (Given that semiprimes are denser than the primes, it is likely that most prime gaps contain a semiprime, but it is highly unlikely (and inconsistent with the prime tuples conjecture) that all of them do.) The "bounded gaps between primes" technology of Goldston-Yildirim-Pintz, Zhang, Maynard, and Polymath does provide prime gaps of short length, but the method also inevitably produces several almost primes in the vicinity of such gaps, and with our current techniques we cannot prevent one of these almost primes being a semiprime inside the prime gap.