Let $D$ be the tangency point of $\odot(I)$ with $BC$ and let $M_A,M_B,M_C$ be the midpoints of $BC,CA,AC,$ resp. $AI$ cuts $BC$ and $M_BM_C$ at $U,V,$ resp and $X \equiv DI \cap M_BM_C.$ Let $T \in OI$ be the orthocenter of $\triangle DEF$ and $Y \equiv DT \cap EF.$ if $I_A$ is the A-excenter of $\triangle {ABC},$ then from $(I,A,U,I_A)=-1$ $\Longrightarrow$ $IV \cdot II_A=IA \cdot IU$ $\Longrightarrow$ $\tfrac{IA}{II_A}=\tfrac{IV}{IU}=\tfrac{IX}{ID}.$ But since $\triangle DEF \cup T$ is similar to the excentral triangle of $\triangle {ABC}$ with orthocenter $I,$ we have $\tfrac{TD}{TY}=\tfrac{II_A}{IA}=\tfrac{ID}{IX}$ $\Longrightarrow$ $XY \parallel \overline{TIO}.$ Now the inversion WRT $\odot(I)$ swaps $Y \longleftrightarrow S$ and $X \longleftrightarrow H$ (this is because it's known that $M_BM_C$ is the polar of $H$ WRT $\odot(I)$) $\Longrightarrow$ $\odot(IHS)$ is tangent to $OI \ (\star).$ Let $A'$ be the projection of $A$ on $OI$ and $P \equiv DI \cap AA'$ $\Longrightarrow$ $ASIA'$ is cyclic on account of its right angles at $S,A'.$ If $M$ is the midpoint of the arc $BC$ of $\odot(ABC),$ we have that $M,D,S$ are collinear (well-known). From $MB^2=MD \cdot MS= MU\cdot MA$ $\Longrightarrow$ $AUDS$ is cyclic $\Longrightarrow$ $\angle SDJ=\angle SAI=\angle SA'I$ $\Longrightarrow$ $SPA'DJ$ is cyclic with circumdiameter $\overline{PJ}$ $\Longrightarrow$ $\angle PSJ=90^{\circ}.$ But from $(\star)$ we get $\angle SHI=\angle SIJ=\angle SAA'$ $\Longrightarrow$ $AHPS$ is cyclic $\Longrightarrow$ $\angle PAH=\angle PSH$ $\Longrightarrow$ $\angle HKA'=90^{\circ}-\angle PAH=90^ {\circ}-\angle PSH=90^{\circ}-(\angle JSH-90^{\circ})=180^{\circ}-\angle JSH$ $\Longrightarrow$ $HKJS$ is cyclic, as desired.