Would you try:
f :: a -> a
f x = undefined :: a
And tell me if it works? IMO it doesn't.
It won't; a will be a new type variable unrelated to the one in the signature, in the absence of ScopedTypeVariables and an explicit forall in the signature.
As I understand it, without ScopedTypeVariables the scope of type variables in the signature is exactly the signature; with ScopedTypeVariables *and* an explicit forall, the scope includes the equation.
Also, I should mention that the extensions-free way to do this kind of thing is judicious use of asTypeOf.