Given $y=f\left(x\right)=x^3-x^2-2x$

$\Rightarrow f\left(1\right)=1-1-2=-2$

$\Rightarrow f\left(-1\right)=-1-1+2=0$

The slope of a line joining the points $\left(x_1, y_1\right) \& \left(x_2, y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$

Thus, the slope of the line segment joining the points $\left(1, f\left(1\right)\right) \& \left(-1, f\left(-1\right)\right)$ is $m=\frac{f\left(1\right)-f\left(-1\right)}{1+1}=\frac{-2-0}{2}=-1$

Given, this line segment is parallel to the tangent of the curve $y=f\left(x\right),$ at $\left(x, y\right)$ and we know that the slope of the tangent to $y=f\left(x\right)$ is $\frac{dy}{dx},$ also the slope of two parallel lines is same, hence 

$m=\frac{dy}{dx}=3x^2-2x-2$

$\Rightarrow 3x^2-2x-2=-1$

$\Rightarrow 3x^2-2x-1=0$

$\Rightarrow \left(x-1\right)\left(3x+1\right)=0$

$\Rightarrow x=1, -\frac{1}{3}.$

Hence, the set $S$ is $\left\{-\frac{1}{3}, 1\right\}.$