Solution:
The given equation is equivalent to
$$a x^{2}+(1-2 a) x+(1-a)=0$$
where $x \neq -1, -\frac{1}{2}, 1$. Hence this equation has two real roots $x_{1}$ and $x_{2}$ such that
$$x_{2}^{2}-a x_{1}=a^{2}-a-1$$
Since $x_{1}+x_{2}=\frac{2 a-1}{a}$ we get that
$$x_{2}^{2}+a x_{2}-a^{2}-a+2=0$$
This together with the identity
$$a x_{2}^{2}+(1-2 a) x_{2}+1-a=0$$
implies that
$$\left(a^{2}+2 a-1\right) x_{2}=a^{3}+a^{2}-3 a+1=\left(a^{2}+2 a-1\right)(a-1)$$
The coefficient of $x_{2}$ vanishes if $a=-1 \pm \sqrt{2}$.
If $a=-1+\sqrt{2}$, then
$$(-1+\sqrt{2}) x_{2}^{2}+(3-2 \sqrt{2}) x_{2}+(2-\sqrt{2})=0$$
which is impossible, since the discriminant of this quadratic equation equals $33-24 \sqrt{2}<0$, i.e. it has no real roots.
If $a=-1-\sqrt{2}$ we get the equation
$$(-1-\sqrt{2}) x_{2}^{2}+(3+2 \sqrt{2}) x_{2}+(2+\sqrt{2})=0$$
that has two real roots, which are not equal to $\pm 1$ and $-\frac{1}{2}$.
Let now $a \neq -1 \pm \sqrt{2}$. Then $x_{2}=a-1$ and hence $a(a-1)(a-3)=0$. Since $a \neq 0,1$ we get $a=3$. In this case the roots of the given equation are $-\frac{1}{3}$ and $2$, and they satisfy the given condition.
Thus the desired values of $a$ are $-1-\sqrt{2}$ and $3$.