As per the post that you have linked, the Duality Theorem for boolean algebra is different:
Let $A'$ be the result of interchanging ∨ and ∧ in $A$, and replacing $P$ by $¬P$ for each atom $P$. Then $A' \Leftrightarrow ¬A$.
The proof is by structural induction.
The above proof does not apply sic et simpliciter to Mendelson's transformation.
We have to note that $P' = P$, for every atom $P$, and neither $P$ nor $P'$ are tautologies. Thus, the result holds trivially.
The same for a formula with a number of nagation whatever, but without occurrences of $\lor$ and $\land$.
Thus, the formula $A$ must contain at least one occurrence of a binary connective and one of negation.
A very simple example (maybe the simplest) of Mendelson's transformation is with formula $A = P \lor \lnot P$.
We have that $\lnot A' = \lnot (P \lor \lnot P)' = \lnot (P' \land \lnot P')=\lnot P \lor P$.
So, in the "basic case" the property holds.
If instead we consider a formula $B=P \land \lnot Q$, which is not a tautology, we have that $\lnot B'=\lnot (P \land \lnot Q)'= \lnot(P' \lor \lnot Q')=\lnot P \land Q$, and neither it is a tautology.
IMO, the result holds considering that for $A$ whatever, we may rewrite it equivalently in CNF, i.e. as a finite conjunction $C_1 \land \ldots \land C_n$ of formulas that are in turn disjunctions: $C_i= D_{i1} \lor \ldots \lor D_{im}$ of atoms or negated atoms.
We have that the CNF is a tautology iff every $C_i$ contains an atom and it negation, i.e. a "basic disjunction" $P \lor \lnot P$.
Consider now $A$ in CNF; what happens with Mendelson's transform? That $A'$ will be in DNF: a disjunction of conjuncts.
What happens when we prefix $\lnot$ to $A'$? That the and's and or's will be exchanged again when we move inside the negation sign and the new formula will be again in CNF.
What happened to the "basic disjunctions"? The first step changes $P \lor \lnot P$ into $P \land \lnot P$, while the second step changes it into $\lnot P \lor \lnot \lnot P$.
Thus, if every disjunct $D_i$ of the CNF corresponding to $A$ contains a "basic disjunction", this will be so also for the corresponding $D'_i$.
Best Answer
The substitution leading to $(s \to r) \leftrightarrow (\neg r \to \neg s)$ is correct, and this formula is a well-known tautology. You are right in saying that it's not enough to consider the case $r = s$; maybe there's an error in your truth table somewhere? It should yield true in every row.