If $a+b+c$ divides the product $abc$, then is $(a,b,c)$ a Pythagorean Triple?

You actually want it the other way around: if $a^2+b^2=c^2$ then $a+b+c|abc$. That you can prove very quickly from the general form of primitive Pythagorean triples $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.

Alternatively: $$\frac{abc}{a+b+c}=\frac{abc(a+b-c)}{(a+b+c)(a+b-c)}=\frac{abc(a+b-c)}{2ab}=\frac{c(a+b-c)}{2},$$ which is a positive integer for both cases: $a,b,c$ are all even; $a,c$ are odd and $b$ is even.