I agree that the link you provide to the proof offered by Proof Wiki is "self-reliant", and seems, to me at least, to invoke the very rule in which we are interested. In a sense, it can be seen as a "model" proof exemplifying the general case of invocation of Disjunction Elimination, $\lor \mathcal E$.
So, yes, this can sort of be thought of as an "axiom": a basic rule of inference we take to be valid, and from which other rules of inference may be derived. It does rely on modus ponens, twice. But the principle that, if, given the truth of $p \lor q$, and some conclusion $r$ follows from $p$, and the same conclusion $r$ follows from $q$, then we can conclude that $r$ is true.
It might be reassuring to note that the argument characterizing "Disjunction Elimination" can be expressed as follows:
$$(((P \to Q) \land (R \to Q)) \land (P \lor R)) \to Q\tag{Tautology}$$
which proves to be tautologically (necessarily) true $(\dagger)$, and so we have very good reason to accept the rule of inference as valid.
$(\dagger)$: Exercise - Confirm, using a truth-table, that the given tautology is, in fact, tautologically true.
Proceed methodically: Suppose the premisses are true and conclusion false. So
$1.\quad p \quad \Rightarrow \quad T$
$2.\quad p \lor q \quad \Rightarrow \quad T$
$3.\quad q \to (r \to s) \quad \Rightarrow \quad T$
$4.\quad t \to r \quad \Rightarrow \quad T$
$5.\quad \neg s \to \neg t \quad \Rightarrow \quad F$
From the last, you know
$6.\quad \neg s \quad \Rightarrow \quad T$
$7.\quad \neg t \quad \Rightarrow \quad F$
Whence
$8.\quad s \quad \Rightarrow \quad F$
$9.\quad t \quad \Rightarrow \quad T$
4 and 9 give us
$10.\quad r \quad \Rightarrow \quad T$
So $r \to s$ is false, and hence, from (3)
$11.\quad q \quad \Rightarrow \quad F$.
So we've worked backwards to successfully find a valuation of all the variables (at lines 1, 8--11) which you can check makes all of 1 to 5 true, i.e. makes the premisses true and conclusion false.
Systematizing this "working backwards" method gives us the user-friendly method of "semantic tableaux" or "truth-trees" used in many textbooks (including mine, and Paul Teller's which is freely available online).
Best Answer
The thing you're missing is the law of non-contradiction:
$$ \neg (P \wedge \neg P) $$
i.e. $\neg c$ when $c$ is a contradiction.
To perform a proof by contradiction -- proving $\neg p$ via a proof of $p \implies c$ -- via proof by contrapositive, let $q$ be $c$.
The form of proof by contradiction you quote follows from the form I mention above by the equivalence $\neg \neg p \equiv p$. (so substitute $\neg p$ into your proof by contrapositive)