Consider the equation $f(x) = 1 \Leftrightarrow x^{4} + a x^{3} + b x^{2} + c x = 1$, since it has four roots $x_{1}, x_{2}, x_{3}, x_{4}$ then we can write it as
$$(x - x_{1})(x - x_{2})(x - x_{3})(x - x_{4}) = 0.$$
Note that $x_{1} + x_{2} = x_{3} + x_{4} = -\frac{a}{2}$ then
$$\left(x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2}\right)\left(x^{2} - (x_{3} + x_{4}) x + x_{3} x_{4}\right) = 0$$
or
$$\left(x^{2} + \frac{a x}{2} + x_{1} x_{2}\right)\left(x^{2} + \frac{a x}{2} + x_{3} x_{4}\right) = 0.$$
Continue with the equation $f(x) = 2 \Leftrightarrow \left(x^{2} + \frac{a x}{2} + x_{1} x_{2}\right)\left(x^{2} + \frac{a x}{2} + x_{3} x_{4}\right) = 1$. By putting $X = x^{2} + \frac{a x}{2}$, we get
$$(X + x_{1} x_{2})(X + x_{3} x_{4}) = 1$$
Since $f(x) = 2$ has four real roots then this equation has two roots like $X = \alpha, X = \beta$. This implies that
$$\left\{ \begin{array}{l} x^{2} + \frac{a x}{2} = \alpha \\ x^{2} + \frac{a x}{2} = \beta \end{array} \right.$$
Then four roots $x'_{1}, x'_{2}, x'_{3}, x'_{4}$ of the $f(x) = 2$ can be divided into two pairs that have the sum equal to $-\frac{a}{2}$ which means $x'_{1} + x'_{2} = x'_{3} + x'_{4}$. This finishes the proof.