Did you mean $N = [\sqrt{n!}]$ in the first line ?

Delaunay's heuristics describe the order of Sha as the curves vary in a family of elliptic curve of fixed rank. In fact, the probability that Sha has a certain order depends on the rank in this model. So these heuristics won't say anything about the rank.

I agree that this works in theory. I am not so sure any computer package can compute the units of $\mathbb{Q}(\mu_p)$ for $p\geq 37$.

Correct, there are no algorithms, but there are very good conjectural algorithms for genus 1 curves, i.e. computer programs that will be able to decide if there is a solution (and find them all). If it can't then some conjecture is wrong (BSD). Of course larger examples, say an intersection of 196 quadrics in $\mathhb{P}^{379}$ won't even have that.

almost all of it. Can you expand ?

I was about to comment that there was no need for Bernoulli numbers, since one can redo the start of the proof of Clausen and von Staudt by hand. And that is exactly what you have outlined here. Since the question was about congruence of $S_m(m)$ modulo $m$ I thought the "$m$-adic development" of it might have some additional interest.

Oh, I see. Maybe, htat works. Here the point $P$ has everywhere good reduction, so the sqrt $e(kP)$ of the denominator of $x(kP)$ satisfies $e(kP) = f_k(P)\cdot e(P)^{k^2} = f_k(P)$ where $f_k$ is the $k$-th division polynomial.

Yop and that seems a good indication to me that the construction of the $p$-adic L-function is going to be more complicated than in the cyclotomic case. The most interesting $p$-adic Lie extension for an elliptic curve is not solvable, all other extensions won't be canonically attached to the curve so the information of the extension needs to enter as well.