Your system has a 1-parameter family of solutions, that is, infinitely many of them. In the solution returned by Solve, the parameter y can be anything (that does not make a denominator zero).

Here's a way to figure out if there's a redundant equation.

vars = {x1, x2, x3, x4, x, y};
sys = {x1 == x A1/(x A1 + y A3), x2 == x A2/(x A2 + y A4), 
   x3 == y A3/(x A1 + y A3), x4 == y A4/(x A2 + y A4), 
   x1 + x2 + x3 + x4 == 2, x + y == 1};
(* calculate rank of system *)
ideal = sys /. Equal -> Subtract;                  (* convert equations to polynomial/rational expressions *)
basis = GroebnerBasis[ideal, vars];
relations = PolynomialReduce[ideal, basis, vars];  (* writes system in terms of basis *)
"rank" -> MatrixRank[relations[[All, 1]]]          (* number of independent equations *)
(* construct minimal system *)
minpos = FirstPosition[#, 1, Nothing] & /@  RowReduce[relations[[All, 1]]]; 
"independent" -> minpos
min = Extract[sys, minpos];                          (* extract the minimal system (drop redundant equation) *)
sol2 = Solve[min, DeleteCases[vars, y]] // Simplify; (* delete y to control which variable are solved for *)
sol = Solve[sys, vars] // Simplify;                  (* Simplify puts solutions in the same form *)
"SameQ[sol1,sol2]" -> sol === sol2                   (* compare solutions *)
(*
  "rank" -> 5
  "independent" -> {{1}, {2}, {3}, {5}, {6}}          <-- equation 4 is redundant
  "SameQ[sol1,sol2]" -> True
*)

Check that the dropped equation is satisfied:

sys[[4]] /. sol // Simplify
(*  {True}  *)

I try now reduce, but it seems still not getting me a solution for say, x variable I am interested in. Reduce[{x1 == x A1/(x A1 + y A3) && x2 == x A2/(x A2 + y A4) && x3 == y A3/(x A1 + y A3) && x4 == y A4/(x A2 + y A4) && x1 + x2 + x3 + x4 == 2 && x + y == 1 && 0 < x < 1 && A1 > 0 && A2 > 0 && A3 > 0 && A4 > 0}, {x1, x2, x3, x4, x, y}, Reals] // Simplify