Newton's Sums with euler's number?

Actually you don't need the second equation.

It is a pretty well-known theorem that if $x + y + z = 0$ , then $x^3 + y^3 + z^3 = 3xyz$ . Therefore, $xyz = \boxed{\frac{e^2}{3}}$

Identity: ${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }-3xyz=\left( x+y+z \right) \left( { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }-xy-yz-zx \right)$

Putting x+y+z=0, ${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }-3xyz=0\\ \Rightarrow { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }=3xyz$

Putting ${ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }={e}^{2}$

${ e }^{ 2 }=3xyz\\ \Rightarrow xyz=\frac { 1 }{ 3 } { e }^{ 2 }$

So $A=1$ , $B=3$ , $C=2$

So, $A+B+C=\boxed { 6 }$

While Ariel's solution is slick, here is the proof of the theorem he invoked.

0=(x+y+z)^3 -3(x^2+y^2 +z^2)(x+y+z)=6xyz -2(x^3 +y^3 + z^3)

So 3xyz=x^3 + y^3 + z^3