Floor equation $x=\frac{⌊x⌋}{x - ⌊x⌋}$

$$x=\frac{⌊x⌋}{x - ⌊x⌋}$$ for $x\in\mathbb R\setminus\mathbb N$.

I've tried putting $x = n + e$, where $n\in\mathbb Z, 0 < e < 1$ and I've got that $n + e = \frac{n}{e}$. Now, I don't know how should I approach problem further, and I am not sure if this is OK what I've done.

Your start is pretty good. Now continue by solving the quadratic equation: $$n+e=\frac ne\iff ne+e^2=n^2\iff e=\frac{\pm\sqrt{n(n+4)}-n}2.$$

The assumption $e\geq0$ implies that $e=\frac{\sqrt{n(n+4)}-n}2$. Now, we rationalize to get

$$\frac{\sqrt{n(n+4)}-n}2=\frac{(\sqrt{n(n+4)}-n)(\sqrt{n(n+4)}+n)}{2(\sqrt{n(n+4)}+n)}=\frac{2n}{\sqrt{n(n+4)}+n}=\frac{2}{1+\sqrt{1+\frac 4n}}<1.$$

Hence, all solutions are of the form $x=n+\frac{2}{1+\sqrt{1+\frac 4n}}$ for $n\in\mathbb N$.