Problems with solutions involving Lambert W function of transcendental equation

Both Reduce and Solve do what they should (see e.g. What is the difference between Reduce and Solve?) even though they might be always better. Reduce may deal with various transcendental functions, however it is not refined enough to work e.g. with elliptic functions, see e.g. 1 and 2. Nevertheless it works well here yielding a bit involved result (we use y instead of l to avoid possible confusion with 1):

Reduce[x  Exp[-2/(x)] == (x - 1) - 1/(2y), x] // TraditionalForm

We can find out that for any $0leq y leq frac{1}{2}$ solution is complex. However one observes that it might work here, although it doesn't

Solve[ x  Exp[-2/(x)] == (x - 1) - 1 /(2y) && (y > 0 && y != 1/2) && x ∈ Reals,
       x, MaxExtraConditions->All]

since the output is expected to be real and unique, see

Plot[ ReIm[1/(-2x Exp[-2/x]+ 2x - 2)], {x, -4, 4}, Evaluated -> True,
      PlotStyle -> Thick, PlotRange-> {-3,3}, AspectRatio -> Automatic,
      Epilog->{ Dashed, Thick, Red, Line[{{1.255,-3},{1.255,3}}],  
                Cyan, Line[{{-4,1/2},{4,1/2}}]}]

On the other hand one might make Solve yield slightly restricted result with Method -> Reduce, e.g.:

sol = FullSimplify[ x/. First @ Solve[ x Exp[-2/(x)] == (x - 1) - 1/(2y), x,
                     Method -> Reduce] /. C[1] -> 0] 

(2 + 4y)/(4y + (1 + 2y) ProductLog[-4y/(E^((4y)/(1 + 2y)) (1 + 2y))])

ay = Limit[ sol, y -> -Infinity]
N @ %

2/(2 + ProductLog[-2/E^2])
1.255

eqn = x Exp[-2/(x)] == (x - 1) - 1/(2 l);

Let x == 2/y && y != 0

eqn2 = eqn /. x -> 2/y // FullSimplify[#, y != 0] &

(* 4 + E^y (-4 + (2 + 1/l) y) == 0 *)

soly = Assuming[l > 0 && y != 0, 
  Solve[eqn2 && y != 0, y, Method -> Reduce][[1]] // Simplify]

Verifying that this satisfies eqn2

eqn2 /. soly // Simplify[#, Element[C[1], Integers]] &

(* True *)

The solution for x is then

solx = (x -> 2/y) /. soly

eqn /. solx // Simplify[#, Element[C[1], Integers]] &

(* True *)

For C[1] == -1 this is the same result as that provided by Roman in his comment.

EDIT:

Real solutions require that C[1] be either -1 or 0

fd = FunctionDomain[{x /. solx, Element[C[1], Integers]}, l] // 
  Simplify[#, Element[C[1], Integers]] &

Solve with the Method->Reduce is actually Reduce.

a) the built-in does not work with the restriction x in Reals. b) the built-in does not work with the restriction l positive. c) the formulation variation between Assumption and &&-logic is just to prefer the recommendation in the documentation of Solve.

ClearAll[x, ll]
Solve[x Exp[-2/(x)] == (x - 1) - 1/(2 ll) && 
  Element[ll, Rationals], x, Method -> Reduce]

---

{{x -> ConditionalExpression[-(I/(2 [Pi] C[1])), 
    C[1] [Element] Integers && ll == -(1/2)]}, {x -> 
   ConditionalExpression[1/(I [Pi] + 2 I [Pi] C[1]), 
    C[1] [Element] Integers && ll == -(1/2)]}, {x -> 
   ConditionalExpression[(2 (1 + 2 ll))/(
    4 ll + ProductLog[C[
      1], -((4 E^(-((4 ll)/(1 + 2 ll))) ll)/(1 + 2 ll))] + 
     2 ll ProductLog[C[
       1], -((4 E^(-((4 ll)/(1 + 2 ll))) ll)/(1 + 2 ll))]), 
    ll [Element] Rationals]}}

And the substitution:

% /. ll -> 15/31

---

{{x -> Undefined}, {x -> Undefined}, {x -> 122/(
   31 (60/31 + 61/31 ProductLog[C[1], -(60/(61 E^(60/61)))]))}}

This result is different. There is a free c1 parameter.

Solve[eq, z, Method -> Reduce]
Solve[eq && z != 0, z, Method -> Reduce]
Solve[eq && z != 0 && Element[z, Reals], z, Method -> Reduce]
Solve[eq && z != 0, z, Reals, Method -> Reduce]
Solve[eq && z != 0 && Element[z, Reals] && Element[f, Reals], z, 
 Method -> Reduce]
Solve[eq && z != 0 && Element[f, Reals], z, Reals, Method -> Reduce]
Solve[eq && z != 0, {z, f}, Reals, Method -> Reduce]
Solve[eq && z != 0, z, Method -> Reduce] /. C[1] -> 0

Solve prefers the condition of the domain in the later argument list and not in the logics of the equations set. Solve prefers more freedom to restrictions. The more general the solution defined in the equations is the better the results. The option Method->Reduce employs actually the built-in Reduce instead of Solve.

I created some inputs for examples. My outputs are different. I use V12.0.0.