YES

We show the termination of R/S, where R is

  f(x) -> x

and S is:

  a -> g(a)

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs


and Q consists of the rules:

  f(x) -> x
  a -> g(a)

No dependency pair remains.