y == ProductLog[x] expansions, suggested by the x E^x term(s)†, pretty much gets you there:

Simplify[x (E^x + 1) Log[x] - x E^x + x^2 + x + 1 /.
  {E^x -> y/x, Log[x] -> Log[y] - x}]
% /. y -> x E^x
(*
-((1 + x) (-1 + y)) + (x + y) Log[y]
-((1 + x) (-1 + E^x x)) + (x + E^x x) Log[E^x x]
*)

†Also suggested by Bill's remark, "The functions (−1+xe ^x ) and (x+lnx) share the same implicit zeros...." That's clear from the (-1+y) and Log[y] factors in each term of the first result.

Assuming you know the desired output form:

x (E^x + 1) Log[x] - x E^x + x^2 + x + 1 /.
  Log[x] -> u - x //
 Collect[#, u, Factor] /. u -> Log[x] + x &
(*  -((1 + x) (-1 + E^x x)) + (1 + E^x) x (x + Log[x])  *)

If I didn't know the desired form, I'd probably investigate by hand, unless I knew from the original problem why one of the forms x + Log[x], x E^x - 1, E^x + 1, etc. was relevant to the problem. Then I would know to put things in terms of it, the way I put the above in terms of u = x + Log[x]. We were fortunate here that Log[x] appears in only one term in the desired form. To replicate the approach with u = x + 1, you'd have to figure out which of the x to substitute — and then make only that substitution. Not easy.