In my book, and everywhere the writter found in internet, to prove that $P(\emptyset)=0$, they do this:
$1=P(Ω)=P(Ω∪∅)=P(Ω)+P(∅) \implies P(∅)=0$ because $Ω∩∅=∅.$
When the writter tried to prove this, she did it differently:
Since $\emptyset \square \emptyset = \emptyset$ (for $\square = \cap, \cup$), we have $P(\emptyset) = P(\emptyset \cup \emptyset) = P(\emptyset)+P(\emptyset)$ and therefore since $(\mathbb{R}, +)$ is a cancellative monoid, it follows that $P(\emptyset) = 0$.
Is this correct? If yes why is the usual proof the former? My proof seems more general as it does not make use of the axiom $P(\Omega) = 1$.
You're correct in that your proof works for any measure space, unnecessarily as it’s an axiom that $\mu(\emptyset)=0$, while the quoted proof works in finite measure spaces. It is slightly more general, yes, but the quoted proofs are in the context of probability spaces so that they only care about finite measure spaces anyway.