Can't this inequality really not be solved with Reduce?

Taking $alpha$ as given, I am trying to understand/know under which restrictions on $n$, the following holds:
$$ 0 leq n+alpha Wleft(-e^{frac{-n}{alpha}} right) leq 1.$$
i.e I want to solve for $n$ the following two inequalities: a) $0 leq n+alpha Wleft(-e^{frac{-n}{alpha}} right)$, and b) $n+alpha Wleft(-e^{frac{-n}{alpha}} right) leq 1$.
In addition, I know $alpha >0$, that both $n$ and $alpha$ are reals.

• I manage to get that a) is satisfied for $ngeq alpha$,
  using: Reduce[n+a*ProductLog[-E^(-n/a)]>=0, n, Reals] which returns (a<0&&n==a)||(a>0&&n≥a).
• However I struggle to find any solution for the inequality b). I have tried all of the following:

(* for convenience I first define the function *)
f[n_, a_] =n+a*ProductLog[-E^(-n/a)]

Reduce[f[n, a]<=1, n, Reals]
Reduce[f[n, a]<=1 && a>0, n, Reals]
Solve[f[n, a]<=1, n, Reals]
Solve[f[n, a]<=1 && a>0, n, Reals]

Which all return:

"This system cannot be solved with the methods available to Solve/Reduce."

Even though I specified the domain as suggested here.

Is the inequality in b) really not solvable for $n$? Could Anyone think of an alternative command or commands that would do the trick?

Thank you,

EDIT:

In my context, f[n_, a_]being strictly increasing in n, and also because I know that (due to the domain of the Lambert W function) we must have n ≥ a (on the graph of f[n, a]below, we see clearly that it is not defined otherwise), even the solution for the equality to 1 would suffice me.

Manipulate[Plot[f[n, a], {n, -1, 10}], {a, 0, 5, 0.05, Appearance -> "Labeled"}]