TL;DR You are correct.
So, in the question you were already given the link to, it is stated that $$f(x) = -\frac x{\ln x}W\left(-\frac{\ln x}x\right),$$
however, it is bit imprecise, because $W$ is multivalued function, with its branches denoted with $W_0$ and $W_{-1}$ and thus, if you were to plot $f$ this way in some software that has Lambert W-function implemented, you would get the same graph as by plotting $x^y-y^x=0.$
Thus, the first step would be to eliminate the line $y=x$ to get your $g$.
We will use identities
\begin{align}
W_0(xe^x) = x,&\ x\geq -1,\\
W_{-1}(xe^x) = x,&\ x\leq -1,
\end{align}
to identify which branch in the above definition of $f$ we need to use:
\begin{align}
f(x) = x &\iff -\frac x{\ln x}W\left(-\frac{\ln x}x\right) = x\\
&\iff W\left(te^t\right) = t,\ x = e^{-t}
\end{align}
By the above identities, we need to use $W_{-1}$ when $t\geq -1$ and $W_{0}$ when $t\leq -1$ to avoid $f(x) = x$. Substituting back with $x$, we get the following definition of $f$:
$$f(x) = \begin{cases}
-\frac x{\ln x}W_{-1}\left(-\frac{\ln x}x\right),\ 0<x< e,\\
-\frac x{\ln x}W_{0}\left(-\frac{\ln x}x\right),\ x\geq e,
\end{cases}$$
but here is the thing, $W_{-1}\left(-\frac{\ln x}x\right)$ is not defined on whole $(0,e)$. The reason is that $W_{-1}$ is only defined on $[-1/e,0)$, so we need to solve $-1/e \leq -\ln x/x < 0$. We get that $x>1$ so the correct definition of $f$ is
$$f(x) = \begin{cases}
-\frac x{\ln x}W_{-1}\left(-\frac{\ln x}x\right),\ \color{blue}{1<x< e},\\
-\frac x{\ln x}W_{0}\left(-\frac{\ln x}x\right),\ x\geq e.
\end{cases}$$
Finally, we can now give precise definition of $g$ as well (using $x^y = e^{y\ln x}$):
$$g(x) = \begin{cases}
\exp\left(-xW_{-1}\left(-\frac{\ln x}x\right)\right),\ 1<x< e,\\
\exp\left(-xW_{0}\left(-\frac{\ln x}x\right)\right),\ x\geq e.
\end{cases}$$
Let us analize what is happening on $(1,e)$. We want to prove that $-xW_{-1}\left(-\frac{\ln x}x\right)$ is strictly decreasing on $(1,e)$. Let $u(x) = W_{-1}\left(-\frac{\ln x}x\right)$. By differentiating $-xu(x)$ we get $$-u(x) + \frac{(\ln x - 1)u(x)}{(1+u(x))\ln x}$$ and we want to prove it is strictly negative on $(1,e)$. Having in mind that $0<\ln x< 1$ and $u(x)\leq -1$ on $(1,e)$, we can manipulate the expression to get
$$-u(x) + \frac{(\ln x - 1)u(x)}{(1+u(x))\ln x} < 0\iff u(x)+\frac1{\ln x} > 0.$$
Now, to see this last thing is true involves some trickery which I will skip. It is not as hard as is very tedious. I can write it down if somebody would be interested in that.
We conclude that $-xu(x)$ is strictly decreasing on $(1,e)$ and so is $g(x)$.
Now, the analysis on $[e,+\infty)$ is much easier. If we denote with $v(x) = W_0\left(-\frac{\ln x}{x}\right)$, then $-1 \leq v(x) < 0$ and $v(x)$ is strictly increasing on $(e,+\infty)$ since both $W_0$ and $-\ln x/x$ are. Calculating derivative of $-xv(x)$ we get
$$-v(x) + \frac{(\ln x - 1)v(x)}{\ln x(1+v(x))}$$
and like the above case, we get that $$-v(x) + \frac{(\ln x - 1)v(x)}{\ln x(1+v(x))}> 0 \iff v(x)\ln x > -1$$ which is true on $(e,+\infty)$ since $v(x)> -1$ and $\ln x> 1$. Thus $-xv(x)$ is strictly increasing on $[e,+\infty)$ and so is $g$.
Having done that, it immediately follows that $g$ has minimum at $x = e$, which is easy to calculate: $g(e) = \exp(-e W_0(-e^{-1})) = \exp(-e\cdot (-1)) = e^e.$
Finally, let us check the limits at boundaries:
$$\lim_{x\to 1^+} g(x) = \exp(\lim_{x\to 1^+} (-xu(x))) = \exp(\lim_{x\to 1^+} (-xW_{-1}(-\frac{\ln x}x))) = +\infty$$ because $\lim_{x\to 0^-}W_{-1}(x) = - \infty.$
To verify that $g(x)\sim x$ at $+\infty$:
$$\lim_{x\to+\infty}\frac{g(x)}{x} = \lim_{x\to+\infty}\frac{\exp(-xv(x))}{x} = \exp(\lim_{x\to+\infty}(-xv(x) - \ln x))$$
and to calculate last limit, we can use Taylor expansion of $W_0$ at $0$:
$$\exp(\lim_{x\to+\infty}(-xv(x) - \ln x)) = \exp(\lim_{x\to+\infty}(-x(-\frac{\ln x}{x}-\frac{\ln^2x}{x^2}+\ldots) - \ln x)) = e^0 = 1.$$
However, $\lim_{x\to +\infty}(g(x)-x) = +\infty$.
