$$\frac{z-i}{2i}$$
$$=\frac{zi-i^2}{2i^2}$$
$$=\frac{zi+1}{-2}$$
$$\frac{(x+iy)i+1}{-2}$$
Since imaginary part is 0 $$\frac{x}{-2}=0$$ So the locus is $x=0$ ie. Y axis.
But the given answer is the line $x=y$.
Can I get an explanation on how that happened?
Your calculations are correct. Namely, setting $x=1$ and $y=1$, we get: $$\Im\left(\cfrac{1+i-i}{2i}\right)=\Im\left(\cfrac{1}{2i}\right)=\Im\left(-\cfrac{i}{2}\right)=-\frac{1}{2}\ne0$$ The answer given is wrong.