Assume you have a filtered probability space $(\Omega, \mathcal{F},\mathcal{F}_t,P)$, assume that $B_t$ is a Brownian motion with respect to this filtered probability space.
Define
$$\text{sign}(x)=1, x \ge 0\\ \text{sign}(x)=-1, x <0.$$
It can then be shown that this stochastic integral is a Brownian motion with respect to $\mathcal{F}_t$:
$$X_t=\int_0^t \text{sign}(B_s)dB_s.$$
Since $X_t$ is also a Brownian motion(this must be shown) we can calculate stochastic integrals with respect to it.
It is stated from various sources that we then have
$$dB_s = \text{sign}(B_s)dX_s.$$
But how do we show this last equation? What is it that allows us to move the $\text{sign}(B_s)$ over? It seems that they have the differential and have divided by $\text{sign}(B_s)$, and used that $1/ \text{sign}(B_s)=\text{sign}(B_s)$. But what allows us to do this?
It's easier to just write out the right side of the equation. By definition of $X$, $dX_s = \text{sign}(B_s)dB_s$, so $$\text{sign}(B_s)dX_s = \text{sign}(B_s)\text{sign}(B_s)dB_s = dB_s$$ because $\text{sign}(x)^2 = 1$ for all $x \in \mathbb{R}$.