YES

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

  g(s(x),y) -> g(f(x,y),y)

and S is:

  f(x,y) -> x
  f(x,y) -> f(x,s(y))

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

  g#(s(x),y) -> g#(f(x,y),y)

and Q consists of the rules:

  g(s(x),y) -> g(f(x,y),y)
  f(x,y) -> x
  f(x,y) -> f(x,s(y))

The weakly monotone algebra (N^3, >_lex) with

  g#_A((x1,y1,z1),(x2,y2,z2)) = (x1, y1, z1)
  s_A((x1,y1,z1)) = (x1 + 2, 2, 1)
  f_A((x1,y1,z1),(x2,y2,z2)) = (x1 + 1, 1, 0)
  g_A((x1,y1,z1),(x2,y2,z2)) = (0, 0, 0)

strictly orients the following dependency pairs:

  g#(s(x),y) -> g#(f(x,y),y)

We remove them from the set of dependency pairs.

No dependency pair remains.