Evaluation of $\sum\limits_{n=1}^\infty \frac{(-1)^n\text{Ei}(n)}{n!}$

Question

Motivation:

This sum came up in a sum of a central gamma function problem:

Evaluation $$\sum_{-\infty}^{-1} \Gamma(n,n)= \pi\left(\frac1e-1\right)i+ \sum_{n=1}^\infty \frac{(-1)^n \text{Ei}(n)}{n!}+\sum_{n=1}^\infty a_n(-e)^n, a_n=\left\{1,\frac38,\frac7{81},\frac{47}{3072},\frac{427}{187500},\frac{13}{43740},\frac{36013}{1037664180},\frac{155011}{42278584320},…\right\} $$

If you change one sign, then you can create an integral representation for a similar, but distinct constant:

$$\sum_{n=1}^\infty\frac{(-1)^n \text{Ei}(-n)}{n!}=\int_1^\infty\frac{1-e^{-e^{-x}}}{x}dx=\lim_{b\to\infty}\left(\ln(b)-\int_1^b \frac{e^{-e^{-x}}}xdx\right)=0.19696002723425413985628823432986472692…$$

Problem:

Here are some attempts at an integral representation or closed form where appears the Digamma function and Exponential Integral function. Here is a proof of the integral representation

$$\sum_\Bbb N \frac{(-1)^n\text{Ei}(n)}{n!}=-\sum_{n=1}^\infty \frac{(-1)^n}{n!}\int_{-n}^\infty \frac{dx}{xe^x}=-\int_0^\infty \lfloor x\rfloor d\left(\frac{(-1)^x\text{Ei}(x)}{x!}\right)= \int_0^\infty\frac{\lfloor x\rfloor(-1)^x \text{Ei}(x)ψ(x+1)}{x!} dx-\pi i\int_0^\infty \frac{\lfloor x\rfloor(-1)^x\text{Ei}(x)}{x!}dx-\int_0^\infty \frac{\lfloor x\rfloor(-e)^x}{xx!}dx=-0.50050747226690385805253965374834233777…≈\frac12$$

One can also try an integral representation like the following using the En function:

$$\text E_1(x)=\int_1^\infty\frac{dt}{te^{tx}}=-\text{Ei}(-x)\implies \text{Ei}(x)\mathop=^?-\int_1^\infty\frac{e^{tx}}{t}dt$$

but the integral diverges:

$$-\int_1^\infty\frac{e^{tx}}{t}dt$$

Where we can actually get the following with a natural extension of the Complementary Bell numbers. Please also refer to the integral representations below:

$$\sum_{n=1}^\infty\frac{(-1)^n\text{Ei(n)}}{n!}=\sum_{n=1}^\infty \frac{(-1)^n\gamma}{n!}+ \sum_{n=1}^\infty \frac{(-1)^n\ln(n)}{n!}+\frac1e\sum_{n=1}^\infty\frac{\mathrm{\tilde B}_n}{nn!} \mathop=^\text{analytic}_\text{continuation}\gamma\left(\frac1e-1\right) + \frac{\mathrm{\tilde B}’_0}{e}+ \frac1e\sum_{n=1}^\infty\frac{\mathrm{\tilde B}_n}{nn!} $$

with the Complementery Bell numbers $\mathrm{\tilde B}_n$ from the simpler Bell Polynomials and @VarunVejalla’s integral representation:

$$\mathrm{\tilde B}_n=\text B_n(-1),\text B_n(x)=e^{-x}\sum_{k=0}^\infty \frac{k^n x^n}{n!}\mathop\implies^\text{analytic}_\text{continuation}\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n!} =\frac{\mathrm{\tilde B}’_0}{e} =e^{-1}\lim_{n\to0}\frac{\mathrm{\tilde B}_n-1}{n}, \frac1e\sum_{n=1}^\infty \frac{\mathrm{\tilde B}_n}{nn!}=\frac1e\int_1^e \ln(\ln(x))e^{-x}dx=\frac1e\int_0^1 \frac{e^{1-e^x}-1}{x}dx$$

Here is another integral representation:

$$f(x)=\sum_{n=1}^\infty \frac{(-1)^n\text{Ei}(nx)}{n!}\implies f’(x)=\sum_{n=1}^\infty \frac{(-1)^ne^{nx}}{xn!}=\frac{e^{-e^x}-1}{x}\implies \sum_{n=1}^\infty \frac{(-1)^n\text{Ei}(n)}{n!}=\int_a^1 \frac{e^{-e^x}-1}{x}dx, \sum_{n=1}^\infty \frac{(-1)^n \text{Ei}(an)}{n!}=0,a≈0.566…$$

How can we find an alternate form of $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n\text{Ei}(n)}{n!}$ ? If there is a closed form, then please include it, but it is optional. Here are some similar problems for help. If you do a double sum, then you can see a relation to the Complementary Bell numbers. Please correct me and give me feedback!