YES
We show the termination of the TRS R:

  *(x,+(y,z)) -> +(*(x,y),*(x,z))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(x,+(y,z)) -> *#(x,y)
p2: *#(x,+(y,z)) -> *#(x,z)

and R consists of:

r1: *(x,+(y,z)) -> +(*(x,y),*(x,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(x,+(y,z)) -> *#(x,y)
p2: *#(x,+(y,z)) -> *#(x,z)

and R consists of:

r1: *(x,+(y,z)) -> +(*(x,y),*(x,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        *#_A(x1,x2) = max{5, x2}
        +_A(x1,x2) = max{7, x1 + 1, x2 + 4}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          *#_A(x1,x2) = 0
          +_A(x1,x2) = max{2, x1 - 4, x2 - 3}
          *#!_A(x1,x2) = 0
          +!_A(x1,x2) = max{1, x1 - 6, x2 - 4}
  
    partial status:
  
      pi(*#) = []
      pi(+) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.