expr = (-1 + (E^(w[1, 0] - 1.55432 w[1, 1]) - E^(-w[1, 0] + 1.55432 w[1, 1]))^2 /
(E^(w[1, 0] - 1.55432 w[1, 1]) + E^(-w[1, 0] + 1.55432 w[1, 1]))^2)

expr /. Power[E, Plus[x_, y__]] :> Inactive[Times][Power[E, x], Power[E, y]]/.
{Power[E, Times[c_, w[i_, j_], ___]] :> Power[T[i, j], Sign[c]],
Power[E, w[i_, j_]] :>T[i, j]} //Activate

For version 9:
expr /. Power[E, Plus[x_, y__]] :> times[Power[E, x], Power[E, y]] /.
{Power[E, Times[c_, w[i_, j_], ___]] :> Power[T[i, j], Sign[c]],
Power[E, w[i_, j_]] :> T[i, j]} /. times -> Times
Update: A slightly more general form allows arbitrary number of terms in the exponent:
f = # /. Power -> power /. power[E, x : PatternSequence[___]] :> (
Times @@ (power[E, #] & /@ x)) /.
{power[E, Times[c_, w[i_, j_], ___]] :> Power[T[i, j], Sign[c]],
power[E, w[i_, j_]] :> T[i, j]} /. power -> Power &
f @ expr
same output as above
expr2 = (-1 + (E^(w[1, 0] - 1.55432 w[1, 1] + 2 w[1, 2]) -
E^(-w[1, 0] + 1.55432 w[1, 1] - 3 w[3, 1]))^2/
(E^(w[1, 0] - 1.55432 w[1, 1]) + E^(-w[1, 0] + 1.55432 w[1, 1]))^2)

f @ expr2

Exp[a] Exp[b]toExp[a + b]. You if you make an intermediate step where you replaceEbyexp(lowercase is important!), then do your replacement (or something like it) withexpin place ofE, it should work. Alternatively, you could wrap the expression withInactivateor something. See my answer here for one possibility. – march May 02 '16 at 20:16