Well-known that $T$ is the mixtilinear touchpoint. Also $D$ lies on the circle $IAB$ (since $SA$ and $SB$ are tangents to $IAB$ we get $SI$ intersects $IAB$ at $D'$ then this is a harmonic quadrilateral, note the other midarc and you're done). Now let $IM$ intersect $IAB$ at $E'$ we have $IA \cdot AE' = E'B \cdot IB$ and also by the harmonic quadruple: $BD \cdot IA = AD \cdot IB$ we get the uniqueness of $E$ and $D$ as the intersections of $IM$ and $IS$ with circle $IAB$ respectively..

Maybe it's better if you don't bash. It could be a better bash if you set (after suitable rotation) $SO$ as the real line. And then you get $b=\overline{a}$ and $s=1$, however, I am not going to continue this bash.

Let $P$ be the midpoint of arc $ACB$ not containing $S$, $X$ be the intersection of $SI$ and $AB$, and $Y \neq M$ be the intersection of circles $SMT$ and $IMD$. It is well-known that $P$ is the center of circle $IAB$, and since $\angle STP = 90^\circ$ and $IT = TD$, then $PI = PD$, which means $IADB$ is cyclic. Now notice that $SX \cdot XT = AX \cdot XB = IX \cdot XD$, so $MX$ is the radical axis of circles $SMT$ and $IMD$, which means $Y$ lies on $AB$. Further, notice that $\angle ITY = \angle SMY = 90^\circ$, so $IY = YD$. This means $\angle IMY = \angle DMY$. Then, $\angle MDE = \angle DMB = \angle IMB = \angle EMB = \angle MED$, so $MD = ME$, and it follows that $AME$ and $BMD$ are congruent. Hence, $AE = BD$.

Nice Config! Let $CI \cap k = F$, by Incenter Excenter Lemma we know that $F$ is the center of $(AIB)$. $$\angle STF = 90^\circ = \angle ITF = \angle FTD \implies \triangle ITF \cong \triangle DTF \ \text{(By S-A-S)}$$Thus $FI = FD \implies D \in (AIB)$. Let $G = MF \cap ED$. $$\angle FMB = 90^\circ = \angle FGD = \angle FGE$$thus $G$ is midpoint of $ED \implies \triangle MGE \cong \triangle MGD \implies \triangle MED$ is isosceles and since $F \in MG \implies FD = FE \implies EABD$ is cyclic. Trapezium $EABD$ is cyclic $\iff$ it is isosceles $\implies AE = BD$. Note that if $EABD$ is not a trapezium then it must be a rectangle and in either case we are done.

Let $D' = (IAB)\cap SI$ and $F = SI\cap AB$. Note that since $IF$ is a symmedian in $\triangle AIB$, we have that $A,I,B,D'$ are harmonic. We have $\frac{\angle A + \angle B}{2} = \angle ATS = \angle AD'T + \angle D'AT \implies \angle D'AT = \frac{\angle A}{2} = \angle BAI$. So $AF$ and $AT$ are isogonal in $\triangle AID'$ but since $AF$ passes through $B$, we have that $T$ is the midpoint of $ID'$. So, $D' = D$ and the rest easily follows. oh @2below thanks edited.

Note that if $N$ is the midpoint of $\overarc{BA}$ , $NT\perp DI\to D\in (BIA)$. Redefine $E$ as the intersection of $(BIA)$ and the line through $D$ parallel to $AB$. Now let $EM\cap (BIA)=I'$, then $$-1=E(BA; M\infty_{AB})=(BA; I'D)\to I'\equiv I.$$q.e.d.

@2above Pitagar wrote: Point $S$ is the midpoint of arc $ACB$ of the circumscribed circle $k$ around triangle $ABC$ with $AC>BC$. *ahem*

My solution looks pretty much same to the above ones, and this is too easy for a #6 in my opinion. Note that triangles $ATI$ and $ITB$ are similar by easy angle chasing. So $AI/BI=AT/IT=AT/TD$, and noting $<AIB=<ATD$ by angle chasing, gives us the similarly between $ATD$ and $AIB$. Thus in triangle $AID$, $AT$ is median, and using $<TAD=<BAI$ by the last similarity, we obtain that $AB$ is symmedian in $AID$. Note that also $<BIT=<IAT=<BAD$, thus $AIBD$ is cyclic with $AB$ - symmedian, and thus harmonic. Therefore, $ID$ is symmedian in $ABI$, so $<AIE=<BID=<BAD=<ADE$, thus $AIDE$ is cyclic, and using that $AIBD$ is cyclic, we easily obtain that $ABDE$ is isosceles trapezoid, which finishes the proof.

With the help of BGR2, the following solution was figured out during JBMO training today, yay! To begin with, let $K$ be the midpoint of $\widehat{AB}$, not containing $C$. Then $KS$ is a diameter, so $\angle KTI = \angle KTS = 90^{\circ}$, which together with $TI = TD$ implies $KI = KD$. On the other hand, $KI = KA = KB$ by the Incenter-Excenter lemma, thus $ADIB$ is cyclic with center $K$. Our aim now is to show that $\angle IKD = 2\angle IED$, as this would imply that $K$ is the circumcenter of $AED$, thus $ADEBI$ would be cyclic (with center $K$) and $AE = BD$ shall follow from the given condition $AB \parallel DE$. Equivalently, we want $\angle IKT = \angle IMA$. Since $\angle KTS = \angle KMA = 90^{\circ}$, it suffices to show $\angle KIM = \angle ISM$, i.e. $KI^2 = KM \cdot KS$. But with $KI = KB$ we reduce to $KB^2 = KM \cdot KS$ and this holds because of the right-angled triangle $KBS$ with altitude $BM$ (more precisely, $\triangle KBM \sim \triangle KSB$).

Firstly I will prove that $\angle ISC= \angle IMB$. Obviously it suffices to prove that $\angle CIS= \angle IMS$. Let $R$ be the contact point of incircle with $AB$ and $X$ be the contact point of the $C-$ excircle with $AB$. Let $K$ be a point such that $IMXK$ is a parallelogram. It is well known that $CX$ is parallel to $IM$, so $C,K,X$ are collinear. Because of the parallelogram we have that $IK=MX=MR$ and $IK$ parallel to $MR$. So $IKMR$ is also a parallelogram and, in fact, a rectangle as $IR \perp \ MR$ . Hence $RK=MI=XK$, so $K$ lies on the perpendicular bisector of $XR$ which is $SM$. So $SCIK$ is cyclic as $\angle K=\angle C=90^\circ$. Hence $\angle CIS=\angle CKS= \angle XKM=\angle KMI= \angle IMS$, as needed. Now back to the problem, let $P$ be the midpoint of the other arc $AB$. Then $ PT \perp \ ID$ and $IT=TD$, hence $PD=PI=PB=PA$, so $AIBD$ is cyclic. Note that $SB, SA$ are tangent to $(AIB)$ and hence $IADB$ is harmonic. So $\angle TBD= \angle IBA=\angle IBC =>\angle IBD=\angle CBT=180- CST=180-\angle CSI=180-\angle IMB=180-\angle IED$, where the last equality holds because $ED$ is parallel to $AB$. Hence $I,B,D,E,A$ are all concyclic and so $AE=BD$, as $AB$ is parallel to $ED$.

We will show that $ABDE$ is an isosceles trapezoid. Let $SM\cap (ABC)=S,N$ and $IM\cap (IAB)=E'.$ Since $\angle NTI=90$ and $TI=TD,$ we have $NI=ND$ which yields $D\in (IAB)$. Also $ME'.MI=MA.MB=MN.MS\implies I,N,E',S$ are cyclic. $SM.SN=SA^2=SI.SD$ hence $I,M,N,D$ are cyclic. \[\angle SDM=\angle IDM=\angle INM=\angle INS=\angle IE'S=\angle ME'S\]\[\angle MDN=\angle MIN=\angle E'IN=\angle NE'I\]\[\angle NDE'=\angle DE'N\]By adding these, we get $\angle SDE'=\angle DE'S\iff SD=SE'$. Also $\angle NSI=\angle NE'I=\angle E'IN=\angle E'SN$. Thus, $SN\perp DE'\iff DE'\parallel AB\implies E'=E$ as desired.$\blacksquare$