$\newcommand{\Arctan}{\operatorname{Arctan}}\newcommand{\Ln}{\operatorname{Ln}}\newcommand{\Arg}{\operatorname{Arg}}\newcommand{\Log}{\operatorname{Log}}$We have $\Log(z^2-z)=\Ln|z^2-z|+i\Arg(z^2-z)$.
$$\begin{align}z^2-z&=(x+iy)^2-(x+iy)\\&=(x^2+y^2-x)+i(2xy-y)\end{align}$$
So $|z^2-z|=\sqrt{(x^2+y^2-x)^2+(2xy-y)^2}$.
And:
$$\begin{align}\Arg(z^2-z)&=\Arctan\left(\frac{\Im(z^2-z)}{\Re(z^2-z)}\right)\\&=\Arctan\left(\frac{2xy-y}{(x^2+y^2-x)}\right)\end{align}$$
Finally:
$$\Log(z^2-z)=\frac{1}{2}\Ln((x^2+y^2-x)^2+(2xy-y)^2)+i\Arctan\left(\frac{2xy-y}{(x^2+y^2-x)}\right)$$
So the questions is how can I define a branch cut of $\Log(z^2-z)$ in the complex plane, and how to draw it?
To find branch cuts it is often easier to think of it through the question: "when is $z^2-z$ in the cut of the parent function" rather than identifying a formula for the composite function $\operatorname{Log}(z^2-z)$ and deducing the branches from there. This is because, for instance, it is easy to make mistakes - your formulas for $\operatorname{Arg}$ are incorrect, as the argument is not always just the arctangent.
The principal logarithm has a cut on: $\Re z\le0,\Im z=0$. We have: $$z^2-z=x^2\color{red}{-y^2}-x+i(2xy-y)$$
The imaginary part is zero iff. $2xy=y$, which has solutions for $y=0$ and $x=1/2$. The real part $x^2-y^2-x$ is nonpositive ($y=0$ case) if $x^2-x\le0$, implying $x\in[0,1]$, and in the $x=1/2$ case when $1/4-y^2-1/2\le0,y^2\ge-1/4$, which is true for all real $y$.
We conclude the cut (viewed in $\Bbb R^2$ for notational clarity) is $[0,1]\times\{0\}\cup\{1/2\}\times\Bbb R$, which gives you the following two lines in the Argand diagram:
What you do need explicit formulae for is determining how the logarithm of $z^2-z$ behaves as you approach (and cross) the above-depicted cut from different quadrants. In particular we can think of $1/2$ as a "very bad" point. But, you must be very careful with your regions and your application of the arctangent formulae for the argument. Also please bear in mind that the answer changes significantly depending on your choice of $\operatorname{Log}$.