Rotate $\triangle{DBC}$ anti-clockwise such that the point $C$ moves to the point $A$. This is possible because $AB=CB$. Denote the location of point $D$ after rotation as $D'$, and connect $D'E$.

Because $CD=AD'$ and $\angle{DBE} = \angle{CBD} + \angle{ABE} = \angle{D'BA} + \angle{ABE}=\angle{D'BE}$, we find $\triangle{EBD} \cong \triangle{EBD'}$ by $SAS$. It follows that $D'E=DE=AE=AD'$, or $\triangle{AD'E}$ is equilateral. Because $AD' = CD = AB = AE$, we find point $A$ is the circumcenter of $\triangle{BED'}$. Hence $$\angle{D'BE}=\frac{1}{2}\angle{D'AE}=\frac{1}{2}\times 60^\circ = 30^\circ\implies \angle{ABC}=\boxed{60^\circ}$$