A

连接$OC$，连接$OB$，

$\because OA$平分$\angle BAC$，$\angle BAC=54^{\circ}$，$AB=AC$，
$\therefore \angle OAB=27^{\circ}$，$\angle ABC=\frac{1}{2}\times \left(180^{\circ}-\angle BAC\right)=63^{\circ}$，
$\because OD$垂直平分$AB$，
$\therefore OA=OB$，
$\therefore \angle OBA=\angle OAB=27^{\circ}$，
$\therefore \angle OBE=36^{\circ}$，
$\because OA$平分$\angle BAC$，$AB=AC$，
$\therefore AO$所在的直线垂直平分$BC$，
$\therefore OB=OC$，
$\therefore \angle OCB=\angle OBE=36^{\circ}$，
根据折叠的性质得，$OE=CE$，
$\therefore \angle ECO=\angle EOC=36^{\circ}$，
$\therefore \angle BEO=\angle ECO+\angle EOC=72^{\circ}$，
$\therefore \angle BOE=180^{\circ}-72^{\circ}-36^{\circ}=72^{\circ}=\angle BEO$，
$\therefore OB=BE$，
$\therefore OC=BE=2$，
$\because \angle OCE=\angle BCO=36^{\circ}$，$\angle COE=\angle CBO=36^{\circ}$，
$\therefore \triangle COE$∽$\triangle CBO$，
$\therefore \frac{CE}{OC}=\frac{OC}{BC}$，
$\therefore \frac{CE}{2}=\frac{2}{2+CE}$，
$\therefore CE=\sqrt{5}-1$或$CE=-\sqrt{5}-1(舍去)$，
故选：$A$.