It's easy to show that $AD\cdot DB = (s-a)(s-b)=\frac{c^2-a^2-b^2+2ab}{4}=\frac{ab}{2}=[ABC]$. $\blacksquare$

Denote our desired product as $P$ If the legs are $a,b$ and the inradius is $r$, note that $$P=(a-r)(b-r)$$We have that $r=\frac{ab}{a+b+\sqrt{a^2+b^2}}$, so $$P=\frac{a^2+a\sqrt{a^2+b^2}}{a+b+\sqrt{a^2+b^2}}\cdot \frac{b^2+b\sqrt{a^2+b^2}}{a+b+\sqrt{a^2+b^2}}$$$$P=\frac{ab(a+\sqrt{a^2+b^2})(b+\sqrt{a^2+b^2})}{(a+b+\sqrt{a^2+b^2})^2}$$$$P=ab\cdot\frac{ab+a^2+b^2+(a+b)\sqrt{a^2+b^2}}{a^2+b^2+2ab+2(a+b)\sqrt{a^2+b^2}+a^2+b^2}$$$$P=\frac{ab}{2}$$

I'll prove something for any triangle $\triangle{ABC}$. Let the incircle touch $BC$ at $D$, $AC$ at $E$, and $AB$ at $F$. Then $$s-a=\frac{AB+AC+BC}{2}-BC=\frac{AF+FB+AE+EC+CD+BD}{2}-BD-CD$$$$=\frac{2AE+2BD+2CD}{2}-\frac{2BD+2CD}{2} = AE$$$$\implies s-a=AE.$$Note that we can derive similar expressions for the remaining of the $5$ segments that have endpoints on a vertex and an adjacent point of tangency. This is also farily well-known so you should probably memorize it or look into/derive similar expressions for lengths on excircles.