Let $\{H\}$ be the orthocenter of $\Delta IBC$ and $P$ be the Circumcenter of $\Delta OM_BM_C$. Then $\measuredangle HIM_A=M_AOP=\pi-\frac{(B+C)}{2}$. Now, $$\begin{cases} \frac{HI}{IM_A}=\frac{2IM_A\sin\frac{A}{2}}{IM_A}=2\sin\frac{A}{2}\\ \frac{OM_A}{OP}=\frac{OM_A}{\frac{OM_A}{2\sin\frac{A}{2}}}=2\sin\frac{A}{2}\end{cases}\implies\frac{HI}{IM_A}=\frac{OM_A}{OP}\implies\Delta HIM_A\stackrel{-}{\sim}\Delta M_AOP \cdots\cdots(\bigstar)$$Now observe that $\overline{IM_A}\parallel\overline{OP}$ and $\overline{HI}\parallel\overline{OM_A}$. Let $\overline{M_AH}\cap\overline{OI}=X$ and let $\overline{M_AP}\cap\overline{OI}=X^*$. Then $\frac{IX}{XO}=\frac{IH}{M_AO}=\frac{IM_A}{OP}=\frac{IX^*}{OX^*}$ from $(\bigstar)\implies X^*\equiv X\implies M_A,P,H$ are collinear. $\blacksquare$

We use complex numbers with $(ABC)$ as the unit circle. Let $x,y,z$ be complex numbers such that $A = x^2, B = y^2, C = z^2, M_A = -yz, M_B = -zx, M_C = -xy.$ Since $I$ is the orthocentre of $\triangle M_A M_B M_C$, $I = -(xy + yz + zx)$. Denote the centroid of $\triangle BIC$ by $G$ and the circumcentre of $\triangle OM_B M_C$ by $O_1$. First, we compute $O_1$. We have that \[ O_1 = \frac{ M_B M_C}{M_B + M_B} = \frac{-xyz}{y+z}. \] By the Incentre-Excentre Lemma, $M_A$ is the circumcentre of $\triangle BIC$. On the other hand \[ G = \frac{B + I + C}{3} = \frac{ y^2 + z^2 - xy -yz - zx}{3} \] It suffices to check that $O_1, M_A, G$ are collinear. Equivalently, $\frac{O_1 - M_A}{G - M_A} \in \mathbb{R}$. First, we simplify this expression. \begin{align*} \frac{O_1 - M_A}{G - M_A} &= \frac{yz - \frac{xyz}{y + z}}{\frac{ y^2 + z^2- xy - yz - zx}{3}+yz} \\ &= \frac{ 3yz(y + z - x)}{(y+z)(y^2 + z^2 + 2yz - xy - xz) } \\ &= \frac{3yz}{(y+z)^2}. \end{align*}Then it is easy to check that \[\overline{ \left( \frac{O_1 - M_A}{G - M_A} \right)} = \frac{\frac{3}{yz}}{(\frac{1}{y} + \frac{1}{z})^2}= \frac{3yz}{y^2 + z^2} = \frac{O_1 - M_A}{G - M_A} \]so we are done.

Let $D$, $E$ and $F$ be the contact points of the incircle with $BC$, $AC$ and $AB$ respectively. Now let $G=BI\cap EF$ and $H=CI\cap EF$. We now get by Iran Lemma that $G,H$ lie on the nine point circle of $BIC$. At the same time notice that $HG\parallel M_BM_C$. So now, by letting $N$ be the midpoint of $H_AI$, where $H_A$ is the orthocenter of $BIC$, we clearly get that $\triangle HNG \sim \triangle M_BOM_C$, so take $K=HM_B\cap M_CG\cap NO$ be its center of negative homothety, which is clearly the incimillienter of the nine point circle of $BIC$ and $M_BOM_C$. Now see how, as $HG\parallel M_BM_C$, then by Ceva, line $IK$ goes through the midpoint of segment $M_BM_C$. But this line is well-known to also go through the midpoint of arc $BAC$. Now, taking the homothety centered at $K$ that sends $N$ to $O$ and $I$ to $S$, we have obtained that it sends $H_A$ to $M_A$. Now notice that $H_AM_A$ is the desired Euler Line, so we're almost done. We now claim that $M_A$ is the excimillicenter of the nine point circle of $BIC$ and $OM_BM_C$. If this is true, then the desired Euler Line goes through the excimillicenter and incimillicenter of the mentioned circumferences, so we'd be done. In order to show this, let $R$ be the center of the nine point circle of $BIC$, $O'$ the center of $OM_BM_C$, and $M$ the midpoint of $BC$. Just notice that $O$ is the reflection of $O$ wrt. $M_BM_C$, so we can compute that $\frac{OO'}{MR}=\frac{OM_A}{OM}$ and that $MR\parallel OO'$ as they are both perpendicular to $M_BM_C$. So $M_A$ is indeed the desired excimillicenter, implying that we are done.