Infinite sum of powerseries likely converges to a powerseries with rational coefficients and has then a simple generating function... proof?

Question

Background: In a discussion in the "tetration-forum" a term $\log(x-L)/L+\log(x-L^*)/L^*$ occured, where $L$ and $L^*$ mean the (complex) primary fixpoint (and its conjugate) of the $\exp()$-function. Because the conjugate values occur symmetrically in the sum of the two formal powerseries, the result is a formal powerseries with real coefficients, and we could as well write $T(x) = 2 \cdot \Re (\log(x - L)/L) $ where $\Re()$ means the real part of its argument.

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Problem: Because the exponential function has infinitely many complex fixpoints I thought what would a formal powerseries $S(x)$ look like, when it is the sum off all such $T()$ - or: when all that fixpoints are taken into account?

Let's write $L_k$ for the $k$'th fixpoint, with $L_1 \approx 0.318131505205 + 1.33723570143 \, î$ , then $$T_k(x)=2 \cdot \Re (\frac {\log(x-L_k)}{L_k}) \tag 1$$ (here, using the convention of W|A, we get the $k$'th fixpoint by $\small {L_k = \exp(-\text{productlog}(-k,-1))}$).

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Now the sum of all that $T_k(x)$ (taken as formal power series) gives another formal powerseries $$ S_n(x) = \sum_{k=1}^n T_k(x) = s_{n:0} + s_{n:1} x + s_{n:2} x^2 + \cdots \tag 2 $$ where -heuristically- it seems, that for $n \to \infty$ the values of $s_{n:0}$ diverges but all other coefficients $s_{n:k}$ converge to a finite value, and moreover, to rational values.

Heuristics: Here are the leading coefficients for some $S_n(x)$ up to $n=2^{16}-1$

  n    s_{n:0}          s_{n:1}          s_{n:2}          s_{n:3}        s_{n:4}            ....
  -------------------------------------------------------------- ----------------------------
    1  -2.44695072007  0.945130773416  0.248253690529  -0.111008639310  -0.0937330420633
    3  -2.93357528303  0.982250645218  0.249909831374  -0.111105628916  -0.0937498749582
    7  -3.29668141804  0.992566300100  0.249991660549  -0.111110673480  -0.0937499978358
   15  -3.63438202949  0.996574875354  0.249999033336  -0.111111067412  -0.0937499999458
   31  -3.96777440601  0.998353868014  0.249999876694  -0.111111106232  -0.0937499999984
   63  -4.30277816467  0.999192862471  0.249999983621  -0.111111110535  -0.0937499999999
  127  -4.64069484715  0.999600341105  0.249999997790  -0.111111111041  -0.0937500000000
  255  -4.98130134424  0.999801140528  0.249999999701  -0.111111111102  -0.0937500000000
  511  -5.32397016248  0.999900812031  0.249999999960  -0.111111111110  -0.0937500000000
 1023  -5.66808038151  0.999950466396  0.249999999995  -0.111111111111  -0.0937500000000
 2047  -6.01314198265  0.999975248290  0.249999999999  -0.111111111111  -0.0937500000000
 4095  -6.35880718549  0.999987627918  0.250000000000  -0.111111111111  -0.0937500000000
 8191  -6.70484432699  0.999993814902  0.250000000000  -0.111111111111  -0.0937500000000
16383  -7.05110555754  0.999996907687  0.250000000000  -0.111111111111  -0.0937500000000
32767  -7.39749940344  0.999998453902  0.250000000000  -0.111111111111  -0.0937500000000
65535  -7.74397059311  0.999999226966  0.250000000000  -0.111111111111  -0.0937500000000
   ....

Conjecture 1: coefficients $s_{n:k}$ are rational in the limit for $n \to \infty$ (except for $s_{n:0}$ which diverges to $-\infty$)

If we assume, that indeed that coefficients converge to rational values, we can arrive at a sequence of integer coefficients by a very simple rational scaling:

$$\begin{array} {} \lim_{n \to \infty} S_n(x)= S(x)&= s_0 &+ 1\cdot x + 1\cdot \frac{x^2}{2!}\frac12 − 2\cdot \frac{x^3}{3!}\frac13 − 9\cdot \frac{x^4}{4!}\frac14 \\ && + 6\cdot \frac{x^5}{5!}\frac15 + 155\cdot \frac{x^6}{6!}\frac16 + 232\cdot \frac{x^7}{7!}\frac17 + ... \end{array} \tag 3$$

Coefficients seem to be known:
The miraculous database of integer-sequences, OEIS, knows this coefficients $[1,1,-2,-9,6,155,232, \ldots]$ saying they have the exponential generating function (which I modify here slightly for my purposes): $$ \begin{array} {} U(x) = \frac{\log(1- x\exp(-x))}x &= -1 &+ 1\cdot x\frac12 + 1\cdot \frac{x^2}{2!}\frac13 − 2\cdot \frac{x^3}{3!}\frac14 − 9\cdot \frac{x^4}{4!}\frac15 \\ &&+ 6\cdot \frac{x^5}{5!}\frac16 + 155\cdot \frac{x^6}{6!}\frac17 + ... \end{array} \tag 4$$

Here the coefficient at the constant term is $u_0=-1$ and is likely different to the value of $s_0$ which is a result of a likely divergent series.

I've a simple modification of the $U()$-function which matches then the conjectured rational coefficients of $S(x)$ even by the indexes:

$$ U_1(x)={\small{\exp(x)-1 \over \exp(x)-x} }\tag {5.a} $$ $$ U_2(x) = \int { \small{\frac{U_1(x)}x}} dx + s_0 \tag {5.b} $$ $\qquad \qquad $ _{$U_1$ is a reformulation of the derivative of $U(x)$ and $U_2(x)$ a termwise integration}

Then Pari/GP gives me the following powerseries:

U_2(x)= s0 + x + 1/4*x^2 - 1/9*x^3 - 3/32*x^4 + 1/100*x^5 + 31/864*x^6 + 29/4410*x^7 - 63/5120*x^8 - 2087/326592*x^9 + 39593/12096000*x^10 + 45973/12196800*x^11 - 146387/522547200*x^12 - 10264123/5782233600*x^13 - 2678759/6258954240*x^14 + 833302651/1225944720000*x^15 + 46063312597/111588212736000*x^16 + O(x^17)

Conjecture 2: The coefficients of the limiting formal powerseries of $S(x)$ are the same as that of $U_2(x)$

This seems to be a nice coincidence - if the assumption (conjecture 1) of convergence of the coefficients in $S(x)$ to that rational values holds. I chewed a bit on how to approach a proof, but didn't have a promising idea yet.

Q1: how could this apparent coincidence of the limit of the sum-of-powerseries in $S(x)$ with the coefficients in $U(x)$ (or better $U_2(x)$) be proved?
Q2: can the value of $s_0$ be expressed by a regularized summation?