We know that:
$$\prod_{n=1}^\infty(e^{-n}+1)=\left(-1,\frac1e\right)_\infty$$ with the Q Pocchhammer symbol $(a,q)_k$, but what if we made it:
$$P=\prod_{n=1}^\infty(ne^{-n}+1)= 2.27396845235252995…?$$
which does not appear in oeis. Here are some other forms of the constant:
$$\prod_{n=1}^\infty(ne^{-n}+1)=\lim_{k\to\infty}\frac{\prod\limits_{n=1}^k (e^n+n)}{e^\frac{k(k+1)}2}=\exp\left(\sum_{n=1}^\infty\ln(ne^{-n}+1)\right)=\exp\left(\lim_{k\to \infty}-\frac{k(k+1)}2+\sum_{n=1}^k \ln(e^n+n) \right)$$
If it helps, multiply over the product’s argument’s inverse’s range using the $-1$st branch of Lambert W:
$$\prod_{n=1}^\infty(ne^{-n}+1) = \prod_{-\text W_{-1}(n-1)=1}^\infty n$$
Using an integral:
$$\sum_{n=1}^\infty\ln(ne^{-n}+1)=-\int_0^\infty \lfloor x\rfloor d(\ln(xe^{-x}+1))=\int_0^\infty \frac{x\lfloor x\rfloor}{e^x+x}-\frac{\lfloor x\rfloor}{e^x+x} dx=0.8215265…$$
Additionally, expanding $\ln(x+1)$ and switching sums gives the polylogarithm function:
$$\ln(P)=-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{n^k e^{-nk}}k=\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\operatorname{Li}_{-n}(e^{-n})$$
$\operatorname{Li}_{-n}(x),n\in\Bbb N$ is a rational function which can be used.
Finally,
OEIS A022629, $a_n$, is about such products and gives $\displaystyle P=\sum_{n=0}^\infty a_ne^{-n}$
Is there a closed form or a new alternate form of the $2.27396845235252995…$ constant in terms of other sums, integrals, special functions, named constants etc?
Time to share one of the most beautiful things I learned on MSE. An approximated value for $\prod_{n\geq 0}\left(1+e^{-n}\right)$ can be derived from Mellin's inversion formula, as Marko Riedel shows here:
$$ \sum_{n\geq 1}\log\left(1-\frac{1}{2^{nx}}\right)\approx -\frac{\pi^2}{6x\log 2}+\log\sqrt{2\pi}-\frac{\log\log 2}{2}-\frac{\log x}{2}+\frac{\log 2}{24} x \tag{1}$$ this approximation is obscenely good. By picking $x=1/\log 2$ $$ \sum_{n\geq 1}\log\left(1-\frac{1}{e^n}\right)\approx -\frac{\pi^2}{6}+\log\sqrt{2\pi}+\frac{1}{24}\tag{2} $$ and by picking $x=2/\log 2$ $$ \sum_{n\geq 1}\log\left(1-\frac{1}{e^{2n}}\right)\approx -\frac{\pi^2}{12}+\log\sqrt{2\pi}-\frac{\log 2}{2}+\frac{1}{12} \tag{3} $$ If we consider the difference between $(3)$ and $(2)$ we get $$ \sum_{n\geq 1}\log\left(1+\frac{1}{e^n}\right)\approx \frac{\pi^2}{12}-\frac{\log 2}{2}+\frac{1}{24}\tag{4} $$ $$ \sum_{n\geq 0}\log\left(1+\frac{1}{e^n}\right) \approx \frac{\pi^2}{12}+\frac{\log 2}{2}+\frac{1}{24}\tag{5} $$ $$ \boxed{\prod_{n\geq 0}\left(1+\frac{1}{e^n}\right)\approx \sqrt{2}\, \exp\left(\frac{\pi^2}{12}+\frac{1}{24}\right)}\tag{6} $$ where the absolute error is less than $9\cdot 10^{-9}$.
I highly doubt there is a nice closed form, but $(6)$ should be more than enough for practical purposes.
Marko Riedel's technique can be adapted to treat $\sum_{n\geq 1}\log\left(1-\frac{n}{2^{nx}}\right)$, too. I do not like to steal credits, so I will leave him the task to fill the missing details.