$\DeclareMathOperator\Im{Im}\DeclareMathOperator\Fix{Fix}$As suggested by Sam, I am posting this as an answer.

As mentioned in the comments, the above question can be solved using a result of Baumeister–Dyer–Stump–Wegener from A note on the transitive Hurwitz action on decompositions of parabolic Coxeter elements, which says the following: let $t_i$, $i=1,\dotsc, k$ be reflections in an arbitrary Coxeter group $(W,S)$. Assume that $W'$ is a (not necessarily standard) parabolic subgroup of $W$, and assume that $t_1 t_2\dotsm t_k$ is a reduced reflection factorization of an element $w\in W'$. Then $t_i\in W'$ for all $i$. Applied to the case $k=2$ and $W'=W_I$ this answers the question.

I also think that there is a more elementary way to see this (for $k=2$). Assume that $w\mathrel{:=}t_1 t_2\in W_I$. Let $V$ be the geometric representation of $W$ and let $\beta_1$ and $\beta_2$ be the roots corresponding to $t_1$, respectively $t_2$. Then $$\Im(t_1 t_2 - 1)\subseteq \mathbb{R} \beta_1 \oplus \mathbb{R} \beta_2.$$ Now since $w\in W_I$, denoting $\alpha_i$, $i\in I$ the simple roots attached to the simple generators in $I$ we also have $$\Im(t_1 t_2 - 1)=\Im(w - 1)\subseteq \bigoplus_{i\in I} \mathbb{R} \alpha_i.$$

**Claim :** $\mathbb{R} \beta_1 \oplus \mathbb{R} \beta_2\subseteq \bigoplus_{i\in I} \mathbb{R}\alpha_i$ (which concludes the proof, as $\beta_i$ are then linear combinations of the simple roots in $I$, hence $t_i\in W_I$).

To see this, by the above inclusions, it is enough to see that $M(w)\mathrel{:=}\Im(w-1)=(w-1)(V)$ has dimension $2$. The point is that, unfortunately, this is *not* the case in general: this is related to what has already been said in the comments but for instance, this fails in type $\widetilde{A}_1$, taking $w=st$, where $\{s,t\}=S$, then the line spanned by $\alpha_s+\alpha_t$ is fixed by $w$, hence $\dim(M(w))=1$. As noted in the comments, as $t_1$ and $t_2$ are distinct, the roots $\beta_1$ and $\beta_2$ are not proportional, and therefore $\Fix(t_1t_2)=\Fix(t_1)\cap\Fix(t_2)$. But these two hyperplanes may be equal, as in the example.

But I think that on can cheat as follows by taking a bigger Coxeter group $\widetilde{W}$ (with geometric representation $V'$) in which $W$ is a standard parabolic subgroup, and such that $(w-1)(V')$ has dimension two. For such an enlarged geometric representation we still have the inclusions $(w-1)(V') \subseteq \mathbb{R}\beta_1\oplus\mathbb{R} \beta_2$ and $(w-1)(V')\subseteq \bigoplus_{i\in I}\mathbb{R} \alpha_i$.

For instance, one can take $V'$ as follows: first, let $x\in W$ such that $x t_1 x^{-1}=s\in S$, and let $w'\mathrel{:=}x w x^{-1}= s t_2'$, with $t_2'=xt_2 x^{-1}$. Consider the Coxeter group $\widetilde{W}$ with one more simple generator $s'$ than $W$, such that $ss'=s's$ and $m_{s't}=\infty$ for any $t\in S\setminus\{s\}$. Let $V'$ be its geometric representation (note that $\dim(V')=\dim(V)+1$). Then $s'$ commutes with $s$ but cannot commute with $t_2'$ (as $s\neq t_2'$), hence $\alpha_{s'}\in\Fix(s)$ but $\alpha_{s'}\notin\Fix(t_2')$, and therefore $\Fix(s)\neq \Fix(t_2')$. It follows that the intersection $\Fix(s)\cap \Fix(t_2')=\Fix(st_2')=\ker(w'-1)$ has dimension $\dim(V')-2$, and hence, that $(w'-1)(V')$ has dimension $2$. Conjugating back by $x^{-1}$ we get that $(w-1)(V')$ has dimension $2$, which concludes the proof.

conjugateto an element that lies in a proper parabolic subgroup. $\endgroup$4more comments