We know that $P \to Q$ is equivalent to $\neg P \lor Q$, as can be verified easily in truth table. Now suppose we have proof for self-implication below [the axiom system is Lukasiewicz's, with L1: $P \to (Q \to P)$, L2: $(P \to (Q \to R) \to ((P \to Q) \to (P \to R))$]:
(1) $P \to ((P \to P) \to P)$ --- (L1)
(2) $(P \to ((P \to P) \to P)) \to ((P \to (P \to P)) \to (P \to P))$ --- (L2)
(3) $(P \to (P \to P)) \to (P \to P)$ --- (1,2 MP)
(4) $P \to (P \to P)$ --- (L1)
(5) $P \to P$ --- (3,4 MP)
This proof establishes that $P \to P$. It is at this point that I realized two things:
First, since we know that $P \to P$ is equivalent to $\neg P \lor P$ and since $\neg P \lor P$ is the Law of Excluded Middle (or more precisely $P \lor \neg P$, but the order of negation don't matter), can it be said that the two are the same? That is, LEM is self-implication. If so, then a logic without LEM (intuitionistic logic) is a logic without self-implication. In other words, in a logic where LEM is not a theorem, self-implication is also not a theorem. Is my reasoning correct?
Second, as a corrolary of the first, can we concludes that the above proof is also (albeit in a disguised form, given semantic reading of equivalence between propositions) proof of LEM?
Self-implication is true in intuitionistic logic as well. It really should be: if we assume $P$, we really should be able to derive $P$.
The point is that $P \to Q$ and $\neg P \lor Q$ are not equivalent in intuitionistic logic. The direction: $P \to Q$ implies $\neg P \lor Q$ uses LEM (and cannot be proved without LEM).
As pointed out in the comments, it may be worth noting that truth tables do not work for intuitionistic logic. That is, if a truth table says two formulas are equivalent, then they are classically equivalent, but not necessarily intuitionistically equivalent. There are similar semantics for intuitionistic logic, called Heyting algebras, but this is more complicated.