Translating $A \to B$ to $\lnot A \lor B$ is, in my opinion, cheating. It's possible to do this directly from the axioms. In fact, you don't even need the third one. Let's call your first axiom K and your second axiom S. We can prove this without the third axiom. Your question amounts to asking the following question in combinatory logic: Can we find an expression in terms of K and S alone for the following lambda expression? $\lambda x . \, \lambda y . \, \lambda z . \, y (x z)$ The answer, by general theory, is yes. Explicitly, $S(K(S(S(K S)K)))K x y z = y (x z)$
Hence our proof takes the following form. For the sake of brevity, we write $A = p \to q$ $B = \lnot q \to \lnot p$
and more generally we write $[n]$ for the $n$-th expression in the list below.
- (Axiom S) $(A \to [11]) \to ([14] \to [15])$
- (Axiom K) $[11] \to (A \to [11])$
- (Axiom S) $[10] \to [11]$
- (Axiom S) $(\lnot q \to [6]) \to ([9] \to [10])$
- (Axiom K) $[6] \to (\lnot q \to [6])$
- (Axiom S) $(p \to (q \to \bot)) \to ((p \to q) \to (p \to \bot))$
- (MP 5, 6) $\lnot q \to [6]$
- (MP 4, 7) $[9] \to [10]$
- (Axiom K) $\lnot q \to (p \to \lnot q)$
- (MP 8, 9) $\lnot q \to (A \to \lnot p)$
- (MP 3, 10) $(\lnot q \to A) \to B$
- (MP 2, 11) $A \to [11]$
- (MP 1, 12) $[14] \to [15]$
- (Axiom K) $A \to (\lnot q \to A)$
- (MP 13, 14) $A \to B$
In full gory detail:
- (Axiom S) $(A \to ((\lnot q \to A) \to B)) \to ((A \to (\lnot q \to A)) \to (A \to B))$
- (Axiom K) $((\lnot q \to A) \to B) \to (A \to ((\lnot q \to A) \to B))$
- (Axiom S) $(\lnot q \to (A \to \lnot p)) \to ((\lnot q \to A) \to B)$
- (Axiom S) $(\lnot q \to ((p \to \lnot q) \to (A \to \lnot p))) \to ((\lnot q \to (p \to \lnot q)) \to (\lnot q \to (A \to \lnot p)))$
- (Axiom K) $((p \to \lnot q) \to (A \to \lnot p)) \to (\lnot q \to ((p \to \lnot q) \to (A \to \lnot p)))$
- (Axiom S) $(p \to \lnot q) \to (A \to \lnot p)$
- (MP 5, 6) $\lnot q \to ((p \to \lnot q) \to (A \to \lnot p))$
- (MP 4, 7) $(\lnot q \to (p \to \lnot q)) \to (\lnot q \to (A \to \lnot p))$
- (Axiom K) $\lnot q \to (p \to \lnot q)$
- (MP 8, 9) $\lnot q \to (A \to \lnot p)$
- (MP 3, 10) $(\lnot q \to A) \to B$
- (MP 2, 11) $A \to ((\lnot q \to A) \to B)$
- (MP 1, 12) $(A \to (\lnot q \to A)) \to (A \to B)$
- (Axiom K) $A \to (\lnot q \to A)$
- (MP 13, 14) $A \to B$
Exercise for enthusiasts. Find a shorter proof, or show that this proof is minimal.