Let $\sigma(x)=\sqrt2|x|$. If $X_t$ is a semimartingale, then $\sigma(X_t)$ is a semimartingale, and \[ \sigma(X_t) = \sigma(X_0) + \sqrt2\int_0^t \text{sgn}(X_s)\,dX_s + 2\sqrt2\Lambda_t(0), \] where $\Lambda_t(0)$ is the semimartingale local time for $X$ at $0$. It follows that the Stratonovich integral of $\sigma(X_t)$ with respect to $B_t$ is well defined, and \[ \int_0^t \sigma(X_s)\circ dB_s = \int_0^t \sigma(X_s)\,dB_s + \frac12[\sigma(X),B]_t, \] where $[\sigma(X),B]_t$ is the cross-variation of $\sigma(X_t)$ and $B_t$. The SDE then becomes \[ dX_t = -X_t\,dt + \sigma(X_t)\,dB_t + \frac12d[\sigma(X),B]_t. \] If $X$ is a solution to this SDE, then the cross-variation is calculated as \begin{align*} [\sigma(X),B]_t &= \sqrt2\int_0^t \text{sgn}(X_s)\,d[X,B]_s\\ &= \sqrt2\int_0^t \text{sgn}(X_s)\sigma(X_s)\,ds\\ &= 2\int_0^t X_s\,ds. \end{align*} Hence, the SDE simplifies to \[ dX_t = \sigma(X_t)\,dB_t, \] and so any solution is at least a local martingale. But since the solutions to this SDE are geometric Brownian motions that we know to be martingales, we are done.
Edit:
Sorry, I just noticed you want $\sigma$ to be strictly positive, but $\sqrt2|x|$ is $0$ at $0$. This can be fixed by taking $\sigma(x)=\sqrt{2(x^2+\varepsilon)}$, where $\varepsilon>0$. In this case, \sigma'\sigma=2x, which is what you want. As an aside, note that as $\varepsilon\to0$, this converges to $\sqrt2|x|$.