I'll use # for the symbol given in the problem.
(I)
a # b = a + b - ab
b # a = b + a - ba
a + b - ab = b + a - ba
So a#b = b#a
True!
(II)
a # 0 = a + 0 - a*0 = a
True!
(III)
(a # b) # c = (a + b - ab) # c = a + b - ab + c - c(a+b-ab) = a + b + c - ab - ac - bc + abc
a # (b # c) = a # (b + c - bc) = a + b + c - bc - a(b+c-bc) = a + b + c - bc - ab - ac + abc
a + b + c - ab - ac - bc + abc = a + b + c - bc - ab - ac + abc
So (a#b)#c = a#(b#c)
True!
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