YES

Problem:
 +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
 *(*(x,y),z) -> *(x,*(y,z))

Proof:
 DP Processor:
  DPs:
   +#(*(x,y),*(a(),y)) -> +#(x,a())
   +#(*(x,y),*(a(),y)) -> *#(+(x,a()),y)
   *#(*(x,y),z) -> *#(y,z)
   *#(*(x,y),z) -> *#(x,*(y,z))
  TRS:
   +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
   *(*(x,y),z) -> *(x,*(y,z))
  LPO Processor:
   argument filtering:
    pi(*) = [0,1]
    pi(a) = []
    pi(+) = 1
    pi(+#) = 1
    pi(*#) = 0
   precedence:
    *# ~ +# ~ + ~ a ~ *
   problem:
    DPs:
     
    TRS:
     +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
     *(*(x,y),z) -> *(x,*(y,z))
   Qed