First, if $A$, $B$, $C$, $D$ are concyclic, by $AB = DE$ and $BC = EA$ we have $\angle BAC = \angle EDA$, $\angle ACB = \angle DAE$, so $\angle ABC = \angle DEA$, which means that $AC = AD$.

Second, if $AC = AD$, let $O$ be the center of the circle ($B$, $C$, $D$, $E$ are on the circle). Then $O$ is on the perpendicular bisector ($AH$) of $CD$. Let $F$ be the symmetric point of $B$ by the line $AH$. Then $F$ is on the circle $O$. $AB \neq EA$, so $DE \neq DF$, and thus $E$, $F$ are not the same points. Now one can see that $\triangle AFD \cong \triangle ABC$, with $AB = DE$, $BC = EA$, and one can get $\triangle AED \cong \triangle CBA$, and so $\triangle AED \cong \triangle DFA$, which indicates $\angle AED = \angle DFA$, and therefore $A$, $E$, $F$, $D$ is concyclic. This means that $A$ is on the circle $O$, and $B$, $C$, $D$ are concyclic.