Given that ‘*’ is an operation that is valid on the set S which consists of all real numbers except 1 i.e., R – {1} defined as a*b = a + b – ab 

Let us assume a + b – ab = 1

⇒ a – ab + b – 1 = 0

⇒ a(1 – b) – 1(1 – b) = 0

⇒ (1 – a)(1 – b) = 0 

⇒ a = 1 or b = 1 

But according to the problem, it is given that a≠1 and b≠1 so, 

a + b + ab≠1, so we can say that the operation ‘*’ defines a binary operation on set ‘S’. 

We know that commutative property is p*q = q*p, where * is a binary operation. 

Let’s check the commutativity of given binary operation:

⇒ a*b = a + b – ab 

⇒ b*a = b + a – ba = a + b – ab 

⇒ b*a = a*b 

∴ Commutative property holds for given binary operation ‘*’ on ‘S’. 

We know that associative property is (p*q)*r = p*(q*r) 

Let’s check the associativity of given binary operation: 

⇒ (a*b)*c = (a + b – ab)*c 

⇒ (a*b)*c = a + b – ab + c – ((a + b – ab)×c) 

⇒ (a*b)*c = a + b + c – ab – ac – bc + abc ...... (1) 

⇒ a*(b*c) = a*(b + c – bc) 

⇒ a*(b*c) = a + b + c – bc – (a×(b + c + bc)) 

⇒ a*(b*c) = a + b + c – ab – bc – ac + abc ...... (2) 

From (1) and (2) we can clearly say that associativity holds for the binary operation ‘*’ on ‘S’.