This answer is unfinished.

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For each nonnegative integer $n$ we have
$$I_n-J_n=\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( \pi p_k(\ln\pi) - k! )-\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)} ( S_n p_k(\ln\pi) - k! )=$$
$$(\pi-S_n)\sum_{k=0}^n \frac{a_k}{\ln^{k+1}(\pi)}p_k(\ln\pi).$$


As far as I see, $P_n(x)$ from the question equals $\widetilde P_n(x)$ from the Wikipedia article. Next, as I see, $e^xp_k(x)=\int_{0}^{x} y^ke^y dy+p_k(0)$ (by the way, in the question is written that $p_k(0)=k!$, whereas Mathcad calculations suggest that $p_k(0)=(-1)^kk!$; for instance, I obtained $\int x^3e^x dx=(x^3-3x^2-6x-6)e^x+C$ and  $\int x^4e^x dx=(x^4-4x^3+12x^2-24x+24)e^x+C$).

Thus we have
$$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(\ln\pi)=$$
$$\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}e^{-\ln\pi}\left(\int_{0}^{\ln\pi} y^ke^y dy+p_k(0)\right)=$$
$$e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}\int_{0}^{\ln\pi} y^ke^y dy+e^{-\ln\pi}\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)=$$
$$\frac 1\pi I_n+\frac 1\pi\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0).$$

Then 
$$I_n-J_n=\left(1-\frac{S_n}{\pi}\right)\left(I_n+\sum_{k=0}^n \frac{a_k}{\ln^{k+1} \pi}p_k(0)\right).$$