I've seen in a couple of places (here and here (top of p.26)) that if S is effectively interpretable in T and T is decidable, then S is decidable. We know that first-order logic in a signature with identity and at least one relation symbol is undecidable (see here). But let T be the theory of linear orders without end points and let S $= \emptyset$ (both in the signature with $=$ and $<$). Then the identity mapping is an effective interpretation of S into T, T is decidable, but S isn't.
What am I missing?