Proving or disproving a basic equation in Boolean algebra

Question

I need to prove or disprove that in any Boolean algebra: if $a+ab=b$ then $a=b=1$ or $a=b=0$.

I build the following truth table: $$ \begin{array}{|c|c|c|} \hline a & b & a+ab \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array} $$ So it does looks like that theorem is true. Can I prove it with algebra? if not, How should I prove it?

Edit: You guys proved it for the binary Boolean algebra. The theorem is for every Boolean algebra (I just gave an example for binary). How can I prove it for every Boolean algebra?

Answer 1

$a + ab = b \iff a(1+b) = b \iff a = b$

Answer 2

Adding $b+1$ to both sides of $a+ab=b$, you get $$ \require{cancel} ab+a+b+1=\underbrace{\cancel{b+b}}_{=2b=0}+1 $$ (recall a Boolean ring has characteristic 2). Thus $(1+a)(1+b)=1$. This only holds if $1+a=1+b=1$, i.e., $a=b=0$.