YES

1 decompositions
 #0 -----------
   1: a() -> b()
   2: a() -> f(a())
   3: b() -> f(f(b()))
   4: f(f(f(b()))) -> b()

@Strongly Commuting
--- R
   1: a() -> b()
   2: a() -> f(a())
   3: b() -> f(f(b()))
   4: f(f(f(b()))) -> b()

--- S
   1: a() -> b()
   2: a() -> f(a())
   3: b() -> f(f(b()))
   4: f(f(f(b()))) -> b()