Comparing the equations we get $x^2(x-1)=y^2(y-1)=z^2(z-1)=K$ for some $K$.Now let $P(t)=t^3-t-K$.Clearly $x,y,z$ are the roots of this equation so $x+y+z=0,xy+yz+zx=-1$ [Pluto1708]: Keeping up with the suggestion: By considering the termwise differences between two equations we have $x^3-x^2=y^3-y^2=z^3-z^2=K$, so $x,y,z$ are the roots of $p(t)=t^3-t^2-K$. This polynomial has three distinct roots iff its discriminant is positive, i.e. iff $K\in\left(-\frac{4}{27},0\right)$. By Viète's formulas we have $x+y+z=1$ and $xy+xz+yz=0$, so $x^2+y^2+z^2=1$. Since $$ x^3 = x^2+K, \qquad y^3=y^2+K,\qquad z^3=z^2+K $$we have $x^3+y^3+z^3 = (x^2+y^2+z^2)+3K = 1+3K$ and $$ 3a = (x^3+y^3+z^3)+2(x^2+y^2+z^2) = 3K+3.$$$a=K+1$ gives $\color{red}{a\in\left(\frac{23}{27},1\right)}$. It is not difficult to check that this is also a sufficient condition for the system to have distinct solutions.

Wellingson wrote: Comparing the equations we get $x^2(x-1)=y^2(y-1)=z^2(z-1)=K$ for some $K$.Now let $P(t)=t^3-t-K$.Clearly $x,y,z$ are the roots of this equation so $x+y+z=0,xy+yz+zx=-1$. Should the polynomial be $t^3 - t^2 - K$?