I think you're on the right track. You can use induction to prove the following:
Lemma
Any statement $\phi$ built up from atomic statements and $\equiv$’s alone is true if and only if it contains an even number of instances of false atomic statements
Proof by structural induction on syntactical formation of $\phi$:
Base: $\phi = A$ for some atomic A. So it would be True iff A is True, i.e. iff $\phi$ contains an even number (0) of instances of false atomic statements. Check!
Step: Consider any statement $\phi \equiv \psi$. Assume the inductive hypothesis holds for $\phi$ and $\psi$. Now: $\phi \equiv \psi$ is true iff either both $\phi$ and $\psi$ are true or both are false iff (inductive hypothses) either both $\phi$ and $\psi$ contains an even number of instances of false atomic statements or both $\phi$ and $\psi$ contain an odd number of instances of false atomic statements iff $\phi \equiv \psi$ contains an even number of instances of false atomic statements. Check!
OK, so now we can prove what you need to prove:
Theorem
Any statement $\phi$ built up from atomic statements and $\equiv$’s alone is a tautology if and only if each atomic proposition that appears in $\phi$ appears an even number of times.
Proof:
'if': If each atomic proposition that appears in $\phi$ appears an even number of times, then $\phi$ contains an even number of instances of false atomic statements, regardless of whether you set any of the atomic propositions to True or False. Hence by the Lemma, $\phi$ will always be true, i.e. $\phi$ is a tautology.
'only if' Proof by contradiction: Suppose $A$ is some atomic statement that occurs an odd number of times in $\phi$. Then if we set $A$ to False, and all other atomic statements in $\phi$ to True, we would end up with an odd number of instances of false atomic statements in $\phi$, so by the Lemma this meas that $\phi$ is False, and hence not a tautology. So, if $\phi$ is a tautology, any atomic proposition that appears in $\phi$ must appear an even number of times.
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
To solve the problem, observe that, for every formulas $\varphi$ and $\psi$, the formulas $$\lnot (\varphi \leftrightarrow \psi) \qquad\qquad \lnot \varphi \leftrightarrow \psi \qquad\qquad \varphi \leftrightarrow \lnot \psi $$ are logically equivalent. It means that in the fragment $\{\lnot, \leftrightarrow\}$ of propositional logic, the truth value of a formula depends on the number of occurrences of $\lnot$, but not on their position. And since the truth values are only two, not even the number of occurrences of $\lnot$ really matters, but only the parity of such a number.
This property gives an intuition about the way to generalize the corollary from the fragment $\{\leftrightarrow\}$ to the fragment $\{\lnot, \leftrightarrow\}$. The generalization is the lemma below.
Lemma. Let $\varphi$ whose only connectives are $\lnot$ and $\leftrightarrow$, and let $v$ be a valuation. Then, $v$ makes $\varphi$ true if and only if the sum of the number of occurrences of $\lnot$ in $\varphi$ and the number of occurrences of statement letters in $\varphi$ made false by $v$ is even.
Proof. By structural induction on the formula $\varphi$. Let us denote by $\#^\lnot_\varphi$ the number of occurrences of $\lnot$ in $\varphi$, and by $\bot^v_\varphi$ the number of occurrences of statement letters in $\varphi$ made false by $v$. We want to prove that $v$ makes $\varphi$ true if and only if $\#^\lnot_\varphi + \bot^v_\varphi$ is even.
Base case. The formula $\varphi$ is a statement letter, hence $\#^\lnot_\varphi = 0$ and either $\bot^v_\varphi = 1$ (if $v$ makes $\varphi$ false) or $\bot^v_\varphi = 0$ (if $v$ makes $\varphi$ true). Clearly, $\varphi$ is true if and only if $\bot^v_\varphi = \#^\lnot_\varphi + \bot^v_\varphi$ is even.
Inductive step. Since $\lnot$ and $\leftrightarrow$ are the only connectives used in the formula $\varphi$, there are only two cases.
$\varphi = \lnot \psi$. By induction hypothesis, $v$ makes $\psi$ true if and only if $\#^\lnot_\psi + \bot^v_\psi$ is even. Clearly, $\#^\lnot_\varphi = \#^\lnot_\psi + 1$ and $\bot^v_\varphi = \bot^v_\psi$. Therefore, $v$ makes $\varphi$ true if and only if $v$ makes $\psi$ false (since $\varphi$ is the negation of $\psi$) if and only if $\#^\lnot_\psi + \bot^v_\psi$ is odd if and only if $\#^\lnot_\varphi + \bot^v_\varphi$ is even.
$\varphi = \psi_1 \leftrightarrow \psi_2$. By induction hypothesis, for each $i \in \{1,2\}$, $v$ makes $\psi_i$ true if and only if $\#^\lnot_{\psi_i} + \bot^v_{\psi_i}$ is even. Clearly, $\#^\lnot_\varphi = \#^\lnot_{\psi_1} + \#^\lnot_{\psi_2}$ and $\bot^v_\varphi = \bot^v_{\psi_1} + \bot^v_{\psi_2}$. Cases:
From the lemma above, it easily follows the syntactic carachterization of tatulogies in the fragment $\{\lnot, \leftrightarrow\}$: a formula $\varphi$ whose only connectives are $\lnot$ and $\leftrightarrow$ is a tautology if and only if $\lnot$ and each statement letter occur an even number of times in $\varphi$.
Proof.
\begin{align} \varphi \text{ is a tautology } &\iff \text{ every valuation } v \text{ makes } \varphi \text{ true } \\ &\overset{\text{by lemma}}{\iff} \text{ for every valuation $v$, the sum of the number of} \\ &\qquad \quad \text{occurrences of $\lnot$ in $\varphi$ and the number of occurrences} \\ &\qquad \quad \text{of statement letters in $\varphi$ made false by $v$ is even } \\ &\iff \lnot \text{ and each statement letter occur an even number of times in $\varphi$.} \end{align} The last equivalence holds because the number of occurrences of $\lnot$ in $\varphi$ does not depend on $v$, and because of the remark below. In particular, the remark below excludes the possibility that the sum of the number of occurrences of $\lnot$ in $\varphi$ and the number of occurrences of statement letters in $\varphi$ made false by $v$ is even because the two numbers are odd. $\qquad\qquad\qquad \qquad\qquad \square$
Remark. Let $\varphi$ be a formula and $p$ a statement letter. Let $v$ be a valuation and $\overline{v}$ be a valuaiton with the same values as $v$ for every statement letter but $p$. The number of occurrences of statement letters in $\varphi$ made false by $v$ and the number of occurrences of statement letters in $\varphi$ made false by $\overline{v}$ have the same parity if and only if the number of occurrences of $p$ in $\varphi$ is even.
Thanks to the lemma above, in the fragment $\{\lnot, \leftrightarrow\}$ you can provide not only a syntactic carachterization of tautologies but also syntactic carachterizations of contradictions (a formula that is always false, in every row of its truth table) and of fomulas that are neither tautologies nor contradictoins. Indeed, let $\varphi$ be a formula whose only connectives are $\lnot$ and $\leftrightarrow$.