Can we simply further the expression? We have the set $\left \{z\in \mathbb{C}\mid z\neq 1, \ \left |\frac{z}{z-1}\right |\leq 1\right \}$ 
Let $z=x+yi$.  
We have that \begin{align*}\frac{z}{z-1}&=\frac{x+yi}{(x-1)+yi}=\frac{\left (x+yi\right )\left ((x-1)-yi\right )}{\left ((x-1)+yi\right )\left ((x-1)-yi\right )}  \\ & =\frac{x(x-1)+y^2-yi}{(x-1)^2+y^2}   =\frac{x(x-1)+y^2}{(x-1)^2+y^2}+\frac{-y}{(x-1)^2+y^2}i\end{align*} 
Then we get \begin{align*}\left |\frac{z}{z-1}\right |&=\sqrt{\left (\frac{x(x-1)+y^2}{(x-1)^2+y^2}\right )^2+\left (\frac{-y}{(x-1)^2+y^2}\right )^2} \\ & =\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4}{[(x-1)^2+y^2]^2}+\frac{y^2}{[(x-1)^2+y^2]^2}} \\ & = \sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}} \end{align*} 
can we simplify that further? 
Is not, we have to solve the inequality from that point. So we have from $\left |\frac{z}{z-1}\right |\leq 1$ the following:
\begin{align*}&\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}}\leq 1 \\ & \Rightarrow 0\leq \frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}\leq 1 \\ & \Rightarrow  0\leq x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq [(x-1)^2+y^2]^2  \end{align*} 
From the second inequality we have \begin{align*}&x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq (x-1)^4+2(x-1)^2y^2+y^4 \\ & \Rightarrow x^2(x-1)^2+2x(x-1)y^2+y^2\leq (x-1)^4+2(x-1)^2y^2\end{align*} How do we continue from here? I got stuck right now. 

We have
$$\left |\frac{z}{z-1}\right |=\frac{|z|}{|z-1|}=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2-2x+1}}\leq 1 \iff x\le \frac12$$
