YES

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

  g(s(x)) -> f(x)
  f(0) -> s(0)
  f(s(x)) -> s(s(g(x)))
  g(0) -> 0

and S is:

  rand(x) -> x
  rand(x) -> rand(s(x))

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

and Q consists of the rules:

  g(s(x)) -> f(x)
  f(0) -> s(0)
  f(s(x)) -> s(s(g(x)))
  g(0) -> 0
  rand(x) -> x
  rand(x) -> rand(s(x))

The weakly monotone algebra (N^2, >_lex) with

  g#_A((x1,y1)) = (x1, y1)
  s_A((x1,y1)) = (x1, y1 + 1)
  f#_A((x1,y1)) = (x1, y1)
  g_A((x1,y1)) = (x1 + 1, y1)
  f_A((x1,y1)) = (x1 + 1, y1 + 1)
  0_A = (1, 1)
  rand_A((x1,y1)) = (x1 + 1, 0)

strictly orients the following dependency pairs:

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

We remove them from the set of dependency pairs.

No dependency pair remains.