Since $x+y=7$, then $x+y+z=7+z$. Thus, the equation $(x+y+z)^{2}=4$ becomes $(7+z)^{2}=4$. Since the square of $7+z$ equals 4, then $7+z=2$ or $7+z=-2$. If $7+z=2$, then $z=-5$. In this case, since $xz=-180$, we get $x=\frac{-180}{-5}=36$ which gives $y=7-x=-29$. If $7+z=-2$, then $z=-9$. In this case, since $xz=-180$, we get $x=\frac{-180}{-9}=20$ which gives $y=7-x=-13$. We can check by direct substitution that $(x, y, z)=(36,-29,-5)$ and $(x, y, z)=(20,-13,-9)$ are both solutions to the original system of equations. Since $S$ is the sum of the possible values of $y$, we get $S=(-29)+(-13)=-42$ and so $-S=42$.