I believe you mean your cases to be $x-y>0$ and $x-y\le 0.$ Still, as you've seen, that's not getting you the results you want.
One alternative is to rearrange the equation to isolate one of the absolute value expressions, then square both sides. This allows you to use the fact that $|u|^2=u^2$ for all real numbers $u.$ You will still have an absolute value expression, so rearrange and square again. Now you have no more absolute value expressions, and can try to determine possible solutions from there. Unfortunately, this is still fairly intractable to solve.
A better alternative is to consider the cases $x+y\ge0$ and $x+y<0.$ In the first case, the equation becomes $$x+y+|x-y|=1.$$ Here's where the slick trick comes in: multiplying both sides of the equation by $\frac12$ gives us $$\max(x,y)=\frac12.$$ Can you see why?
In the second case, the equation becomes $$-x-y+|x-y|=1,$$ or equivalently, $$x+y-|x-y|=-1.$$ Again, we multiply by $\frac12,$ giving us $$\min(x,y)=-\frac12.$$ Do you see why?
Putting the cases together, the solution set is a square, which can be described by $\max\bigl(|x|,|y|\bigr)=\frac12.$
Let me know if you have any questions.