Multiplying both sides by $(x-1)(m-1)m$ gives us
\[xm(x-1)-m(m+1)=x(x-1)(m-1)\]
\[x^2m-xm-m^2-m=x^2m-xm-x^2+x\]
\[-m^2-m=-x^2+x\]
\[x^2-x-(m^2+m)=0\]
The roots of this quadratic are equal if and only if its discriminant
($b^2-4ac$) evaluates to 0.
This means
\[(-1)^2-4(-m^2-m)=0\]
\[4m^2+4m+1=0\]
\[(2m+1)^2=0\]
\[m=-\frac{1}{2} \Rightarrow \boxed{\textbf{(E)}}\]
In the equation $\frac {x(x - 1) - (m + 1)}{(x - 1)(m - 1)} = \frac {x}{m}$ the roots are equal when 
$\textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\frac{1}{2}\qquad\textbf{(C)}\ m = 0\qquad\textbf{(D)}\ m =-1\qquad\textbf{(E)}\ m =-\frac{1}{2}$