a = b               <--a and b can stand for the same quantity. 
a² = ab             <--we can multiply both sides by a.  
a² - b² = ab - b2   <--we can subtract b² from both sides.
(a-b)(a+b) = b(a-b) <--we factor both sides of the equation.
a+b = b             <--we divide both side by (a-b)
b+b = b             <--since a = b, we can substitute b for a
2b = b              <--combine like terms b+b and get 2b
2 = 1               <--we divide both sides by b 
                 

The fallacy is in the step colored red. Since a=b, (a-b)=0 and
we may never divide by 0, even when it's camouflaged as (a-b).

Edwin

a = b
 = 
 -  =  - 
(a-b)(a+b) = b(a-b)     <<< You can not cancel the term (a-b), because it is zero due to a = b (see very first line). You can not divide by zero!!! Here is the key! 
                            The rest of the proof is just not relevant.  It is very old math joke.
a+b = b
b+b = b
2b = b
2 = 1