NO

Problem:
 a() -> b()
 f(g(a())) -> f(a())

Proof:
 Nonconfluence Processor:
  terms: f(g(b())) *<- f(g(a())) ->* f(a())
  Qed
  
  first automaton:
   final states: {1}
   transitions:
    f(3) -> 1*
    b() -> 2*
    g(2) -> 3*
  
  second automaton:
   final states: {4}
   transitions:
    f(5) -> 4*
    a() -> 5*
    b() -> 5*