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, >) with

  g#_A(x1,x2) = x1
  s_A(x1) = x1 + 1
  f_A(x1,x2) = x1
  g_A(x1,x2) = x1

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.