Apologies for the unspecific title.
Correct me if I have not understood this correctly, which is why I've posted this question:
$B$ is a Boolean algebra. Prove for $x, y \in B$ that $x \cdot y' = 0$ if and only if $x \cdot y = x$
Can this be proven with identity law? For $x \cdot y = x$ to be true, $y$ needs to be $1$ (considering identity law). That makes $x \cdot y' = 0$.
Is this valid proof?
You're misinterpreting the identity law: it says that if $y=1$ then $x\cdot y=x$, but it doesn't say the converse, which is what you're using.
I recommend thinking about some specific Boolean algebras and playing around the operations, to get a sense of what they're acting like. You'll quickly stumble across nontrivial examples where $x\cdot y=x$ but $y\ne1$.
As for your actual question: consider $x\cdot(y+y')$....