Questions
Is the asymptotic relationship correct? How do I determine $c_1$ and $\kappa$? As, $|s| \to 0$
$$ \sum_{r=1}^\infty s^r \ln(r) \sim c_1 \sqrt{s} + (\kappa - 1 + \frac{\ln(2 \pi)}{2} )s$$
Is it possible to find $A_r$ using this relationship (for any of the r's)? Where,
$$ \sum_{r=1}^\infty A_r(s) \ln(p_r) \sim c_1 \sqrt{s} + (\kappa - 1 + \frac{\ln(2 \pi)}{2} )s$$
Background
Recently I found a technique as in a previous question: Number-theoretic asymptotic looks false but is true?
This led me to another series with interesting properties:
$$ K(s) = s \ln 1 + s^2 \ln 2 + s^3 \ln 3 + \dots = \sum_{r=1}^\infty s^r \ln(r) $$
$$ \implies K(s) = A_1(s) \ln 2 + A_2(s) \ln(3) + A_3(s) \ln 5 + \dots = \sum_{r=1}^\infty A_r(s) \ln(p_r) $$
Where $p_r$ is the r'th prime
Now consider the following:
$$ K(s) = \sum_{r=1}^\infty s^r \ln(r) $$ $$ \implies \sum_{r=0}^\infty s^r K(s) = s \ln 1! + s^2 \ln 2! + s^3 \ln 3! + \dots $$ $$ \implies \frac{ K(s)}{1-s} = \sum_{r=1}^\infty s^r \ln r! $$
Using Stirling approximation:
$$ \implies \frac{ K(s)}{1-s} = \sum_{r=1}^\infty s^r (r \ln(r) - r +\frac{1}{2}\ln(2 \pi) + \frac{1}{2} \ln r + \sum_{k=1}^\infty a_k/r^k) $$
where $a_n$ is the coefficients of the $1/r$ error terms
$$ \implies \frac{ K(s)}{1-s} = s \frac{dK}{ds} - \frac{s}{(1-s)^2} + \frac{\ln(2 \pi) s }{2(1-s)} + \frac{K}{2} + a_1 \text{Li}_1(s) + a_2 \text{Li}_2(s)\dots $$
where $\text{Li}_r(s)$ is the $r$'th poly-logarithmic function. Taking $|s| \to 0 $ and using the asymptotic relation of $\text{Li}_r(s) \sim s$
$$ \implies \frac{ K(s)}{2} \sim s \frac{dK}{ds} - s + \frac{\ln(2 \pi) s }{2} + \kappa s $$
whre $\kappa= \sum a_r$ and now solving the differential equation:
$$ K(s) \sim c_1 \sqrt{s} + (\kappa - 1 + \frac{\ln(2 \pi)}{2} )s$$
The series
$$ \sum_{r=1}^{\infty} s^r \ln(r) = s^2 \ln(2) + s^3 \ln(3) + \cdots $$
is, in fact, an asymptotic expansion with respect to the asymptotic scale $\{s^r\}_{r\in\mathbb N}$ as $s \to 0$. In particular, it is its own asymptotic expansion as $s \to 0$. Proving this would be a very valuable exercise if it is not immediately obvious.
One particular consequence of this fact is that
$$ \sum_{r=1}^{\infty} s^r \ln(r) = s^2 \ln(2) + O(s^3) \qquad \text{as } s \to 0, \tag{$*$} $$
so it can't possibly be true that an asymptotic for the series includes a term involving $\sqrt{s}$ since that would violate $(*)$. No terms which decay more slowly than $s^2$ may appear in an asymptotic for the series unless they are cancelled out by some other term.
Now, to try to address your proposed derivation:
I'm not an expert in the polylogs, but you might want to be more careful "solving" an "asymptotic differential equation". Making something like that rigorous is difficult because there are so many cases where doing that gives you the wrong answer. Assuming what you've done prior to that step is correct, that must be where your error lies. You have found an example where "solving" an "asymptotic differential equation" gives you nonsense.