$a(a'+b) = aa'+a'b = a'b$ ($aa' = 0$ in any case)
$a+a'b = 1a + a(a'+b) = a(1+a'+b) = a$
$a+ab = a(a+b) = a => a+a'b = a+ab$
However, when I use truth table to compare the result of $a+a'b$ to $a+ab$
when $a = 0, b = 1$
$a+ab = 0$ while $a+a'b = 1$.
What is the mistake I made here?
On
$$a + a'b = a + ab = a$$ This is not correct. So $$a + ab = a$$ $$a + ab = a(1 + b) = a$$ This is correct. But $$a + a'b ≠ a$$ The a' is redundant. $$a + a'b = a(1 + b) + a'b = a + ab + a'b = a + b (a + a') = a + b$$ $$a + a'b = a + b$$ $$a + ab = a ≠ a + a'b$$
Also: In your first step you do $a(a'+b) = aa'+a'b = a'b$. This is correct, but I do not understand where $a(a'+b)$ came from. It is not in your starting question. You can multiply a term by 1 because X' + X = 1, but you cannot multiply a term by whatever. The rules have to satisfy boolean algebra.
Was this ( $a(a'+b)$ ) in the original question? Clarify and someone can help.
I think your mistake is at the very first step, where you wrote $$a(a'+b)=aa'+a'b.$$ That should be $$a(a'+b)=aa'+ab.$$