This should hold for any pair of isogonal conjugates. The key claim is that we have $\triangle ABM \sim \triangle ACN$. Since $\angle BAM=\angle CAN$ it suffices to show $\frac{AB}{AM}=\frac{AC}{AN}$. By ratio lemma on the two triangles, we have $$\frac{AB\sin \angle BAM'}{AM\sin \angle MAM'}=\frac{AC\sin \angle CAN'}{AN\sin \angle NAN'}=1 \implies \frac{AB}{AM}=\frac{\sin \angle MAM'}{\sin \angle BAM'}=\frac{\sin \angle NAN'}{\sin \angle CAN'}=\frac{AC}{AN},$$proving the claim. To finish, note that this means $\measuredangle M'MN'=\measuredangle BMA=\measuredangle ANC=\measuredangle M'NN'$. $\blacksquare$

Yes, it holds for any pair of isogonal conjugates. The easiest way I think (which is overkill I guess) is to just characterize those lines. As it says "midpoints" then just make a harmonic pencil. Consider $P, Q$ a pair of isogonal conjugates instead of just $H, O$. Let $X, Y$ be the second intersections of $AP, AQ$ with $(ABC)$. Since they're isogonal we have $XY \| BC$ meaning that if $\ell := \text{perpendicular bisector of } BC$ then $X, Y$ are reflections over $\ell$. Now let's take some point $Z$ satisfying that $(B, Y; X, Z) = -1$. By symetry, if $W := \text{reflection of } Z \text{ over } \ell$ then we have $(C, X; Y, W) = -1$. Hence $(AB, AQ; AP, AZ) = -1 = (AC, AP; AQ, AW)$ so we have $BM \| AZ$ as well as $CN \| AW$. Looking at $(ABC)$, the arcs $ZX$ and $YW$ are equal so $\angle(AZ, AP) = \angle(AQ, AW)$. By the previous parallelism, we obtain $\angle(BM, AP) = \angle(AQ, CN)$ and we're done!

It is well known that $AH$ and $AO$ are isogonal conjugates with respect to angle $A$. Let the reflections of $B$, $M$, and $M'$ about the angle $A$ bisector be $B_1$, $M_1$, and $M'_1$. We know that $B_1$ lies on $AC$, $M_1$ lies on $AH$ and $M'_1$ lies on $AO$. By homothety we have ${B_1}{M_1}{M'_1}||{C}{N}{N'}$. Thus ${M_1}{M'_1}$ and ${N}{N'}$ are parallel and ${M}{M'}$ and ${N}{N'}$ are antiparallel with respect to angle $A$. Thus $MM'NN'$ are concyclic.

Attachments:

We prove that the problem is true for any pair of isogonal conjugates. Swap $H$ , $O$ with $P$ , $Q$ and denote the midpoint of $AB$,$AC$ in order as $X$,$Y$ We get that $XM'\|AN'$ and $AM'\|YN'$ therefore from $\measuredangle XM'A = \measuredangle AN'Y$ and $\measuredangle XAM' = \measuredangle YAN'$ we conclude that $\triangle M'AX \sim \triangle N'AY$ hence $\triangle M'AB \sim \triangle N'AC$ and therefore $\measuredangle AM'B =\measuredangle NM'M =\measuredangle AN'C =\measuredangle MN'N$ hence $M , M' , N , N'$ lie on a circle.

In fact, there's an even more general problem. Quote: Let $ABC$ be a triangle and let $\ell _B$ and $\ell _C$ be two lines through $A$ that are isogonal with respect to $\angle BAC$. A line through $B$ cuts $\ell _C$ at $M$ and $\ell _B$ at $M'$. A line through $C$ cuts $\ell _B$ at $N$ and $\ell _C$ at $N'$ such that $\frac{BM'}{BM}=\frac{CN'}{CN}$. Prove that $M,M',N,N'$ are concyclic. Let $r=\frac{BM'}{BM}=\frac{CN'}{CN}$. Then if $\ell _C'$ is the image of $\ell _C$ under the homothety with centre $B$ and ratio $r$ and $\ell _B'$ is the image of $\ell _B$ under the homothety with centre $C$ and ratio $r$, we have $M'=\ell _C'\cap BM$ and $N'=\ell _B'\cap CN$. Consider the homothety with centre $A$ and ratio $\frac{AC}{AB}$ followed by the reflection wrt the angle bisector of $\angle BAC$. It maps $B$ to $C$ and $\ell _B$ to $\ell _C$, so it maps $\ell _C'$ to $\ell _B'$, therefore it maps $M'$ to $N'$, and hence it maps $BM'$ to $CN'$. This gives that $M,M',N,N'$ are concyclic, as desired.