I am wondering how I may obtain a closed form for $s$ defined as:
$$s=\ln(e+\cfrac1{\ln(e+\cfrac1{\ln(e+\cfrac{1}{\ddots})})})$$
We have:
$$s = \ln\left(e+\frac1s\right) \implies e^s=e+\frac1s \implies e^ss-es-1=0$$
But I am unsure of how we could find a closed form for the root of the RHS equation if we instead consider $f(x)=e^xx-ex-1$. I nervously tag this question lambert-w because of that $e^xx$ term, although I have never seen the function in action and am not sure how to use it in solving this problem.
You are looking for the zero of function $$f(s)= e^ss-es-1$$ which, as said, does not show analytic solution (I think that the solution could be given in terms of the generalized Lambert function).
But $$f'(s)=e^s (s+1)-e$$ cancels at $$s_*=W\left(e^2\right)-1$$ and the second derivative test shows that this is a minumum.
Make a Taylor expansion around $s_*$ to get $$f(s)=f(s_*)+\frac 12 (s-s_*)^2 f''(s_*)$$ which then gives as an approximation $$s_{(2)} \sim s_* + \sqrt{-2\frac {f(s_*)}{f''(s_*)}}$$ that is to say $$s_{(2)} \sim t-1+ \sqrt{\frac 2 e \frac{ e t^2-(2 e-1) t+ e}{(t+1)}} \quad \text{where} \quad t=W(e^2)$$ Numerically, this gives $s=1.388$ while the exact solution, obtained using Newton method, is $1.257$. This is not very good and I think that Newton would be better.
Inverse symbolic calculators do not find anything looking like $1.25675061853776723178$ which is the exact solution.
However, if we expand for one more term, we have a cubic equation in $(s-s_*)$. Using the hyperbolic method, the new approximation is given by $$s_{(3)} \sim \frac{2 (t+1) \cosh \left(\frac{1}{3} \cosh ^{-1}(P)\right)+t^2-3}{t+2}$$ where $$P=\frac{3 e t^4+(3+5 e) t^3-12(e-1) t^2-3(5 e-4) t+11 e}{e (t+1)^3}\quad \text{and} \quad t=W(e^2)$$ This give $s_{(3)}=1.27684$
Edit
All the above was done to show Lambert function appearing in the expression. However, without using Newton method, we can make successive approximations of the solution building the $[1,n]$ Padé approximant of $f(s)$ around $s=1$. This should give $$\left( \begin{array}{ccc} 0 & \frac{1+e}{e} & 1.36788 \\ 1 & \frac{5+2 e}{3+2 e} & 1.23706 \\ 2 & \frac{26+48 e+12 e^2}{8+36 e+12 e^2} & 1.26022 \\ 3 & \frac{126+948 e+792 e^2+144 e^3}{30+516 e+648 e^2+144 e^3} & 1.25615 \\ 4 & \frac{744+17280 e+38160 e^2+20160 e^3+2880 e^4}{144+6960 e+25200 e^2+17280 e^3+2880 e^4} & 1.25685 \\ 5 & \frac{5160+309840 e+1620000 e^2+1915200 e^3+734400 e^4+86400 e^5}{840+101040 e+864000 e^2+1396800 e^3+648000 e^4+86400 e^5} & 1.25673 \\ 6 & \frac{41040+5866560 e+64804320 e^2+154224000 e^3+120960000 e^4+36288000 e^5+3628800 e^6}{5760+1622880 e+28516320 e^2+95558400 e^3+93744000 e^4+32659200 e^5+3628800 e^6} & 1.25675 \end{array} \right)$$