Problem with `Derivative`/`Series` and `InverseFunction`

The problem with using an anonymous pure function inside of InverseFunction is that you get nested anonymous pure functions when evaluating the second derivative. Compare the following 2 TracePrint outputs:

TracePrint[InverseFunction[Function[x, Log[x]+Exp[x]]]'', Derivative[1][_]];
TracePrint[InverseFunction[Log[#]+Exp[#]&]'', Derivative[1][_]];

((1/(Function[x,Log[x]+Exp[x]]^[Prime])[InverseFunction[Function[x,Log[x]+Exp[x]]][#1]]&)^[Prime])

((1/((Log[#1]+Exp[#1]&)^[Prime])[InverseFunction[Log[#1]+Exp[#1]&][#1]]&)^[Prime])

Notice how the the first trace output has a mixture of x and #1, while the second trace output replaces x with #1. The latter trace is an example of a nested pure function using anonymous pure functions, which is why it doesn't work. The issue of nested anonymous pure functions has come up before, see:

1. Is anonymous pure function a scoping construct?
2. Pure function inside another pure function

Addendum

You also ask:

So, what's the best way to get the Series of that (and other similar) function?

Instead of taking the series of the inverse function, you can apply InverseSeries to the series expansion of the function. For example:

InverseSeries @ Series[Log[x]+Exp[x], {x, 1, 2}] //TeXForm

$1+\frac{x-e}{1+e}-\frac{(e-1) (x-e)^2}{2 (1+e)^3}+O\left((x-e)^3\right)$

Series[
    InverseFunction[Function[x, Log[x]+Exp[x]]][x],
    {x, E, 2}
] //TeXForm

$1+\frac{x-e}{1+e}+\frac{(1-e) (x-e)^2}{2 (1+e)^3}+O\left((x-e)^3\right)$

For clarity, I used an expansion point of 1 for the function, which corresponds to an expansion point of E for the inverse function. To get the expansion of the inverse function around 1 you could do:

pt = x /. First @ Solve[Log[x] + Exp[x]==1 && 0<x<1, x];
Simplify[
    InverseSeries @ Series[Log[x] + Exp[x], {x, a, 1}],
    Exp[a]+Log[a]==1
] /. a->pt //TeXForm

$\operatorname{Root}\left[\left\{e^{\#1}+\log (\#1)-1\&,0.51222243303322994816075176697026116734`20.304437036448245\right\}\right]+\frac{x-1}{e^{\operatorname{Root}\left[\left\{e^{\#1}+\log (\#1)-1\&,0.51222243303322994816075176697026116734`20.304437036448245\right\}\right]}+\frac{1}{\operatorname{Root}\left[\left\{e ^{\#1}+\log (\#1)-1\&,0.51222243303322994816075176697026116734`20.304437036448245\right\}\right]}}+O\left((x-1)^2\right)$

which yields the same result:

Series[
    InverseFunction[Function[x, Log[x]+Exp[x]]][x],
    {x, 1, 1}
] //TeXForm

$\operatorname{Root}\left[\left\{e^{\#1}+\log (\#1)-1\&,0.5122224330332299481607867201842576827690521506497275993993`30.\right\}\right]+\frac{x-1}{e^{\operatorname{Root}\left [\left\{e^{\#1}+\log (\#1)-1\&,0.5122224330332299481607867201842576827690521506497275993993`30.\right\}\right]}+\frac{1}{\operatorname{Root}\left[\left\{e^{\#1}+\log (\#1)-1\&,0.5122224330332299481607867201842576827690521506497275993993`30.\right\}\right]}}+O\left((x-1)^2\right)$

InverseFunction and Derivative[1] operate on functions. So, supply them with functions:

h=.
f = InverseFunction[h];
f'[x]

Derivative[1][h][InverseFunction[h][x]]^(-1)

In your case, this leads to:

h = x \[Function] Log[x] + Exp[x];
f'[x]

1/(E^InverseFunction[Function[x, Log[x] + Exp[x]]][x] + 1/ InverseFunction[Function[x, Log[x] + Exp[x]]][x])

Now also Series works. I use machine precision numbers in order to make the output readible:

Series[f[x], {x, 1., 4}] // TeXForm

$$0.512222+0.276146 (x-1.)+0.0225572 (x-1.)^2-0.0123554 (x-1.)^3-0.0000787926 (x-1.)^4+O\left((x-1.)^5\right)$$