Let $b,\sigma:\mathbb R\to\mathbb R$ be Lipschitz continuous and at most of linear growth. Moreover, let $a:=\sigma^2$ and $$s:=\int_0^{\;\cdot\;}\exp\left(-2\int_0^y\frac ba(x)\:{\rm d}x\right){\rm d}y.$$ Assume $s(\mathbb R)=\mathbb R$ (are there nice sufficient conditions on $b,\sigma$ that ensure this?). Now, let $g:\mathbb R\to[0,\infty)$ be Borel measurable with $$\int g(x)\:{\rm d}x=1.\tag1$$
If $b\in C^1(\mathbb R)$ and $\sigma\in C^2(\mathbb R)$, then $$L^\ast h:=\frac12(ah)''-(bh)'\;\;\;\text{for }h\in C^2(\mathbb R).$$ is well-defined. Assuming $g\in C^2(\mathbb R)$, it's straightforward to deduce that $$L^\ast g=0\Leftrightarrow g=\frac ca\exp\left(2\int_0^{\;\cdot\;}\frac ba(x)\:{\rm d}x\right)\tag2.$$ However, the expression found for $g$ would even be well-defined without the assumptions $b\in C^1(\mathbb R)$, $\sigma\in C^2(\mathbb R)$ and $g\in C^2(\mathbb R)$. We simply need to note that linear growth assumptions yield local integrability of $a\sigma$ and $b\sigma$. Further, we need to assume that $b/a$ is locally integrable. $L^\ast g=0$ has then to be understood in the distributional sense. Is it possible to show $(2)$ in that setting too?