As you want to suggest in your comment, you can demonstrate the equaivalence with explicit examples to which it applies, but then this is merely a motivation to define "if-then" in terms of "not-or".
If you start said definition of "if-then", then the semantical explanation 1) shows that you can replace "if-then" with "not-or".
You ask for other elaborations, but if you don't let the semantic explanation count, then to show that equivalence, you must provide the allowed modes of reasoning. This is no quibble, as validity of the rule ((p -> q) -> (¬p v q)) itself is the matter of discourse here.
If we don't want to use semantic explanation, nor formal syntactic deduction for how else "if-then" can be understood, then we'd have to rely on a natural explanation - but you will not be able to find one now, as the "real" reading of "if-then" reasoning was and is a matter of heated philosophical debate. Relevance logic is one approach to it. But p => q always being equivalent to (¬p v q) is not even true in intuitionistic logic. The mathematical structures called Heyting algebras provide a whole family of related formal interpretation of the => connective.
Lastly, I want to comment on your explanation 2). To me it's a little ambigous how to interpret what you want it to say, but either it's a weaker semantical argument than 1) or you say you want to use it to "show" the equivalence deductively. Let's take a look with a classical eye, but without using what we want to show: 2) says ((¬p v q) => (p -> q)) => (¬p v q). Let's investigate the premise (¬p v q) => (p -> q). First level: Assume its premise (¬p v q) and then consider (p -> q). Second level: Assume its premise p. Then classically ¬(¬p) and so by the first level assumption q, so (p -> q) is true. So (¬p v q) => (p -> q) is true anyways and you showed "true => (¬p v q)". From this (¬p v q) actually follows but now the critique of relevance fully applies: The conclusion (¬p v q) wasn't needed in the proof and so you haven't shown much.