Inequality found with the help of Desmos and WA:
Let $x>0$ sufficiently large then we have :
$$x^{\frac{x}{x+1}}-\left(\frac{x}{x+1}\right)^{x}+\ln\left(x\right)<x<x^{\frac{x}{x+1}}-\left(\frac{x}{x+1}\right)^{x}+\ln\left(x\right)+\frac{1}{e}$$
The Puiseux Series for $x\to \infty$ is :
$$f(x)=x^{\frac{x}{x+1}}-\left(\frac{x}{x+1}\right)^{x}+\ln\left(x\right)=x-\frac{1}{e}+ \frac{((\ln^2(x))0.5+\ln(x)-\frac{1}{2e}}{x}+O\left(\left(\frac{1}{x}\right)^2\right)$$
So I think that it shows the RHS .
For the LHS I have tried to differentiate .I don't expose the derivative here because it's ugly and not really useful.
Edit some thought:
We have :
$$f(x)=x^{\frac{x}{x+1}}+\ln(x)+e^{-1}\frac{\ln(x)}{\ln(x)+1}$$
And :
$$f''(x)=\frac{\frac{\left(x^{\frac{x}{x+1}+1}\right)((x + 1)^2+x\log^2(x))}{(x + 1)^4}-1}{x^2}-e^{-1}\frac{(\ln(x)+3)}{x^2(\ln(x)+1)^3}$$
And for sufficiently large $x$ : $f''(x)<0$
To show it we can replace the $x^{\frac{x}{x+1}+1}$ by $x^2$.The rest is not hard.
Let $g(x)$ be :
$$g(x)=-\left(\frac{x}{x+1}\right)^{x}$$
We have :
$$g''(x)=-\frac{\left(\frac{x}{x+1}\right)^{x-1}\left((x(x +1)\ln^2\left(\frac{x}{x+1}\right) + 2x\ln\left(\frac{x}{x+1}\right)+1\right)}{(x+1)^2}<0$$
So $x-(f(x)+g(x))$ is decreasing for sufficiently large $x$ and the limit is 0 .
Question :
How to show the LHS ? Is my edit correct ? Have you an alternative proof?
I tried to make a bit more for this interesting problem.
With $$f(x)=x^{\frac{x}{x+1}}-\left(\frac{x}{x+1}\right)^{x}+\ln\left(x\right)$$ what you want to show is that, for $x \geq 14$ $$f(x) < x < f(x)+\frac 1 e$$ that is to say that $$\lim_{x\to \infty } \, (x-f(x))=\frac 1 e-\epsilon \quad \text{and} \quad \lim_{x\to \infty } \, (f(x)-x+\frac 1 e)=\epsilon$$ with $\epsilon >0$.
Developed as series $$f(x)=x-\frac 1 e+\frac 1 {2e}\sum_{n=1}^\infty (-1)^n \frac {a_n- e tP_{n}(t)} {x^n}\quad \text{where}\quad t=\log(x)$$ Where the $a_n$ are the unknown sequence $$\left\{1,\frac{5}{12},\frac{5}{24},\frac{337}{2880},\frac{137}{1920},\frac{67177}{1 451520},\frac{18289}{580608},\cdots\right\}$$ and the first polynomials are $$\left( \begin{array}{cc} n & P_n(t) \\ 1 & t+2 \\ 2 & \frac{t^2}{3}+2 t+2 \\ 3 & \frac{t^3}{12}+t^2+3 t+2 \\ 4 & \frac{t^4}{60}+\frac{t^3}{3}+2 t^2+4 t+2 \\ 5 & \frac{t^5}{360}+\frac{t^4}{12}+\frac{5 t^3}{6}+\frac{10 t^2}{3}+5 t+2 \\ 6 & \frac{t^6}{2520}+\frac{t^5}{60}+\frac{t^4}{4}+\frac{5 t^3}{3}+5 t^2+6 t+2 \end{array} \right)$$ where interesting patterns can be found.
Limited to the very first term, we have $$f(x)=x-\frac{1}{e}+\frac{e \log (x) (\log (x)+2)-1}{2 e x}+O\left(\frac{1}{x^2}\right)$$ $$f'(x)=1+\frac{-\log ^2(x)+\frac{1}{e}+2}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ $f'(x)$ is always positive and asymptotic to $1^-$.
At this point, I am stuck from a formal point of view.
Waiting for some inspiration, a few numerical results for $x=e^k$ (you probably already obtained similar things) $$\left( \begin{array}{ccc} k & e^k-f(e^k) & f(e^k)-e^k+\frac 1e \\ 3 & 0.04058733 & 0.32729211 \\ 4 & 0.16130501 & 0.20657443 \\ 5 & 0.25346965 & 0.11440979 \\ 6 & 0.30932075 & 0.05855869 \\ 7 & 0.33941666 & 0.02846278 \\ 8 & 0.35454032 & 0.01333912 \\ 9 & 0.36179657 & 0.00608287 \\ 10 & 0.36516437 & 0.00271507 \\ 11 & 0.36668844 & 0.00119100 \\ 12 & 0.36736447 & 0.00051497 \\ 13 & 0.36765948 & 0.00021996 \\ 14 & 0.36778646 & 0.00009298 \\ 15 & 0.36784049 & 0.00003895 \end{array} \right)$$ while $\frac 1e=0.36787944$.
Edit
Concerning the new part of the question, using the new notations, the asymptotics is $$x-(f(x)+g(x))=\frac{1}{e \log (ex)}+\frac{1-e \log (x) (\log (x)+2)}{2 e x}+O\left(\frac{1}{x^2}\right)$$