The solutions of
$$X^3+aX^2+bX+c=0$$
have sum $-a$ and product $-c$. Since $b$ can still vary, your conjecture is definitely not true with complex or real numbers instead of naturals. On the other hand, the same argment with a quadratic polynomial shows that the case with $2$ numbers is true even if we allow real or complex numbers.
For naturals start with an arbitrary almost-solution (e.g. found by pushing around prime factors) such as $1\cdot 8\cdot 9=3\cdot 4\cdot 6$ but $1+8+9\ne 3+4+6$. Then try to find $r,s$ such that $1r+8+9s=3r+4+6s$, i.e. $3s=2r-4$, say $s=2, r=5$.
This gives us a solution $(1r,8,9s,3r,3,6s)$, i.e.
$$ 5\cdot 8\cdot 18 = 15\cdot 4\cdot 12\qquad \text{and}\qquad 5+8+18=15+4+12.$$