18 Pentagons!

We notice at the center of the 18-gon, six pentagons are joined together by their angle that corresponds to $\angle A \implies \angle A= \frac{360^\circ}{6}=60^\circ$ .

Since all the sides of the pentagon are equal, $\triangle ABE$ is equilateral and quadrilateral $BCDE$ is a rhombus.

$\angle ABC$ is an interior angle of the 18-gon, so $\angle B= \angle ABC= 160^\circ$ .

Then $\angle EBC=\angle ABC-\angle ABE= 160^\circ-60^\circ = 100^\circ.$

which implies $\angle D= \angle CDE= \angle EBC= 100^\circ$ and $\angle C= \angle BED=180^\circ -\angle EBC= 180^\circ - 100^\circ =80^\circ$ .

Finally, $\angle E= \angle AED = \angle AEB+\angle BED= 60^\circ+ 80^\circ=140^\circ.$

To summarize: $\color{#D61F06}\angle A= 60^\circ,\color{#20A900} \angle B= 160^\circ,\color{#3D99F6} \angle C=80^\circ,\color{#69047E}\angle D= 100^\circ,\color{#EC7300}\angle E=140^\circ$

$\implies S= 60^\circ+160^\circ+80^\circ+100^\circ+140^\circ=\boxed{540^\circ}.$

For the Bonus question, the answer is yes they are collinear but requires to prove $\angle XYZ= 180^\circ$ . (I will provide the proof later)