Let $\tau>0$, $E$ be a $\mathbb R$-Banach space and $v:[0,\tau]\times E\to E$ such that $v(\;\cdot\;,x)$ is continuous for all $x\in E$ and $v$ is Lipschitz continuous with respect to the second argument uniformly with respect to the first. This is enough to ensure that there is a unique $X^{s,\:x}\in C^0([s,\tau],E)$ with $$X^{s,\:x}(t)=x+\int_s^tv\left(r,X^{s,\:x}(r)\right)\:{\rm d}r\;\;\;\text{for }t\in[s,\tau]\tag1$$ for all $(s,x)\in[0,\tau]\times E$. Now let $$T_{s,\:t}:=X^{s,\:x}(t)\;\;\;\text{for }x\in E$$ for $0\le s\le t\le\tau$.
Let $0\le r\le s\le t\le\tau$. I want to show that $$T_{s,\:t}\circ T_{r,\:s}=T_{r,\:t}\tag2.$$
I guess we need to argue by uniqueness, by I'm not sure how exactly. Let $x\in E$. We may clearly write $$T_{r,\:s}\left(T_{s,\:t}(x)\right)=x+\int_r^sv\left(q,T_{q,\:s}\left(T_{s,\:t}(x)\right)\right)+\int_s^tv\left(q,T_{q,\:t}(x)\right)\:{\rm d}q\tag3,$$ but I don't see how to conclude from here.
The desired claim is clearly trivial and most probably that's why I'm uncertain whether I made a mistake in the following reasoning:
The claim should follow from the following result: Let $0\le r\le s\le\tau$ and $x\in E$, then $$\left.X^{r,\:x}\right|_{[s,\:\tau]}=X^{s,\:X^{r,\:x}(s)}\tag4.$$ We should be able to prove this in the following way: Let $y:=X^{r,\:x}(s)$. Then, $$X^{r,\:x}(t)=y+\int_s^tv\left(q,X^{r,\:x}(q)\right)\:{\rm d}q\;\;\;\text{or all }t\in[r,\tau]\tag5$$ and $$X^{s,\:y}(t)=y+\int_s^tv\left(q,X^{s,\:y}(q)\right)\:{\rm d}q\;\;\;\text{for all }t\in[s,\tau]\tag6$$ and hence $$X^{r,\:x}(t)-X^{s,\:y}(t)=\int_s^tv\left(q,X^{r,\:x}(q)\right)-v\left(q,X^{s,\:y}(q)\right)\:{\rm d}q\;\;\;\text{for all }t\in[s,\tau]\tag7.$$ Let $c$ denote the Lipschitz constant of $v$. Then, $$\left\|X^{r,\:x}(t)-X^{s,\:y}(t)\right\|_E\le c\int_s^t\left\|X^{r,\:x}(q)-X^{s,\:y}(q)\right\|_E\:{\rm d}q\tag8.$$ Thus, by Gronwall's inequality$^1$, $$\left\|X^{r,\:x}(t)-X^{s,\:y}(t)\right\|_E\le0\cdot e^{c(t-s)}=0\tag9\;\;\;\text{for all }t\in[s,\tau].$$ Am I missing something? Does Gronwall's inequality hold for $c_1=0$? I don't see any step in its proof which should not work in that case. So, it should hold in that case as well.
Gronwall's inequality: If $f\in C^0([a,b])$ with $$f(t)\le c_1+\int_a^bc_2(s)f(s)\:{\rm d}s\;\;\;\text{for all }t\in[a,b]\tag{10}$$ for some $c_1\in\mathbb R$ and nonnegative $c_2\in C^0([a,b])$, then $$f(t)\le c_1\exp\left(\int_a^bc_2(s)\:{\rm d}s\right)\;\;\;\text{for all }t\in[a,b].\tag{11}$$