Let $M$ be the midpoint of the side $AB$. Then $\angle OBC = \angle OCB = 90^\circ - \angle A$ and $MX$ parallel to $BC$ comes to $\angle MXB = 90^\circ - \angle A$. But $\angle B'XB = \angle XOC = 2\angle OBC = 180^\circ - 2\angle A$. In the triangle $ABB'$ we have $MA = MB = MB'$ and $\angle BMB' = \angle MAB' + \angle MB'A = 2\angle A$. It follows that the quadrilateral $MBXB'$ is cyclic, hence $\angle MXB = \angle MB'B = \angle ABB' = \angle OBC$, which leads to the conclusion.

Second solution. Let $K$ the projection of $B$ onto the parallel through $A$ to the line $BC$. The quadrilateral $AKBB'$ is cyclic, hence $\angle AB'K = \angle ABK = 90^\circ - \angle B = \angle OCA = \angle CB'X$. It follows that points $K, B'$ and $X$ are collinear. Finally, the triangle $BXK$ is isosceles, with $BX = KX$, therefore $X$ lies on the perpendicular bisector of the line segment $BK$, i.e. $X$ is on the midline of triangle $ABC$ that is parallel to $BC$. We have $\angle XBK = 90^\circ - \angle OBC = \angle A = \angle XKB$ (from the cyclic quadrilateral $AKBB'$). The conclusion follows readily.

Third solution. (Given by Alexandru Mihalcu.) Let $S$ be the midpoint of the line segment $AC$ and $T$ be the intersection point of lines $BX$ and $AC$. As $OS$ is parallel to $BB'$, we have $\frac{B'S}{ST} = \frac{OB}{OT} = \frac{OC}{OT} = \frac{B'X}{XT}$. From the converse of the Angle Bisector Theorem it follows that $XS$ is the angle bisector of $\angle B'XT$. Then $\angle SXT = \frac{\angle B'XT}{2} = \frac{\angle XOC}{2} = \angle OBC$, and the conclusion follows.