jmc

Assume $k\ne 0.$ Then $$x+2=x(kx-1)=k{x}^2-x,$$so $k{x}^2-2x-2=0.$ This quadratic has exactly one solution if its discriminant is 0, or $(-2{)}^2-4(k)(-2)=4+8k=0.$ Then $k=-\frac{1}{2}.$ But then $$-\frac{1}{2}{x}^2-2x-2=0,$$or $x^2+4x+4=(x+2{)}^2=0,$ which means $x=-2,$ and $$\frac{x+2}{kx-1}=\frac{x+2}{-\frac{1}{2}x-1}$$is not defined for $x=-2.$

So, we must have $k=0.$ For $k=0,$ the equation is $$\frac{x+2}{-1}=x,$$which yields $x=-1.$ Hence, $k=\boxed{0}$ is the value we seek.