The given quadrilateral is an irregular shape. Let's transform it to a regular shape by rotating $\triangle{AEB}$ around point $A$ for $90\circ$ counterclockwise.

Because $AB=AD$, we know point $B$ will move to $D$ after such rotation. Denote the new location of $E$ as $E'$. Now $$\angle{BAD}=90^\circ \implies \angle{EAE'}=90^\circ$$ and $$\angle{BAD}=\angle{DCB}=90^\circ \implies \angle{B}+\angle{ADC}=180^\circ \implies \angle{ADE'}+\angle{ADC}= 180^\circ$$This means points $E'$, $D$, and $C$ are collinear. Because $AE=AE'$, it is obvious now that $AECE'$ is a square whose area equals 1. Therefore$$S_{ABCD}=S_{AECE'}=\boxed{1}$$