YES

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

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

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