Can we say that the conclusion in this argument: (P v Q), P |- Q breaks "The Law of Excluded Middle"? And that is the reason why argument is invalid?
I recently studied "The Law of Excluded Middle":
In logic, the law of excluded middle (or the principle of excluded middle) states that for any proposition, either that proposition is true or its negation is true. Wiki
$$\begin{array}{|c|c|c|} \hline p&q&p∨ q\\ \hline T&T&T\\ T&F&T\\ F&T&T\\ F&F&F\\\hline \end{array}$$
"The following argument: (P v Q), P |- Q is invalid. Because,
Premise 1: there are three instance in truth table where (P v Q) is True (1st three in above table),
Premise 2: there are two instance in truth table where (P) is True for (P v Q) to be true at the same time (1st two in above table),
Conclusion: In this scenario Q is both True and False for (P) and (P v Q) to be true, right? and that is the reason why this argument is invalid.
The law of the excluded middle has nothing to do with why this argument is false. The law of the excluded middle says that $P \lor \lnot P$ is always true, but even in logics that do not have that law, you cannot conclude $Q$ from $P$ and $P \lor Q$.
Using your truth table interpretation of $\vdash$, the reason $P, P \lor Q \vdash Q$ is false is that there is a line in the truth table where $P$ and $P \lor Q$ are true, but $Q$ is false, namely, the second one.