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

  r1

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            *#_A(0,0) = 0
            *#_A(0,1) = 0
            *#_A(1,0) = 0
            *#_A(1,1) = 0
            +_A(0,0) = 0
            +_A(0,1) = 0
            +_A(1,0) = 0
            +_A(1,1) = 0
            *_A(0,0) = 0
            *_A(0,1) = 1
            *_A(1,0) = 0
            *_A(1,1) = 1
            *#!_A(0,0) = 0
            *#!_A(0,1) = 0
            *#!_A(1,0) = 0
            *#!_A(1,1) = 0
            +!_A(0,0) = 0
            +!_A(0,1) = 0
            +!_A(1,0) = 0
            +!_A(1,1) = 0
            *!_A(0,0) = 0
            *!_A(0,1) = 0
            *!_A(1,0) = 0
            *!_A(1,1) = 0
  
    partial status:
  
      pi(*#) = [2]
      pi(+) = [1, 2]
      pi(*) = [2]

The following rules are removed:

  p1, p2

Then no dependency pair remains.