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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              *#_A(x1,x2) = x2
              +_A(x1,x2) = x1 + x2 + 1
              *#!_A(x1,x2) = x2 + 1
              +!_A(x1,x2) = x1 + x2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            *#_A(x1,x2) = 0
            +_A(x1,x2) = max{2, x1 - 2, x2 - 2}
            *#!_A(x1,x2) = max{2, x1 - 1, x2 - 8}
            +!_A(x1,x2) = max{1, x1 - 7, x2 - 12}
        
  
    partial status:
  
      pi(*#) = []
      pi(+) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.