jmc

We draw the pentagon as follows, and draw altitude $\overline{BG}$ from $B$ to $\overline{AE}$. Since $\angle BAG=45^{\circ }$, $AG=GB$. We extend lines $BC$ and $ED$ past points $C$ and $D$ respectively until they intersect at $F$. $△CFD$ is a 45-45-90 triangle with $CF=FD=\frac{2}{\sqrt{2}}=\sqrt{2}$. So $GBFE$ is a square with side length $2+\sqrt{2}$, and $AG=BG=2+\sqrt{2}$. It follows that $AE=AG+GE=2(2+\sqrt{2})=4+2\sqrt{2}$, and finally $a+b=\boxed{6}$.