Let $u:[0,t_{e}]\to\mathcal{D}(A)$ satisfy
$$\begin{cases} \frac{du}{dt}=Au & 0\le t \le t_{e} \\ u(0)=x \end{cases}$$
I want to prove that necessarily $u(t)=T(t)x$.
So it's clear to see that $T(0)x=x=u(0)$. And $\frac{d}{dt}T(t)x=AT(t)x=Au$, if $u(t)=T(t)x$.
How can I complete this proof? I assume, at least, by making the supposition that $u(t)\ne T(t)x$ and then looking for a proof by contradiction.
You proved that $T(t)x$ is a solution. So, it remains to show that the problem has only one solution.
Let $u$ and $v$ be two solutions. Notice that the function $w=u-v$ satisfies $$\left\{\begin{align} &\frac{dw}{dt}=Aw, & 0\le t \le t_{e} \\ &w(0)=0. \end{align}\right.$$ Pick $s\in[0,t_e]$ and define $\phi:[0,s]\to X$ be setting $\phi(t)=T(t)w(s-t)$. Then, $$\begin{align} \frac{d}{dt}\phi(t)&=\frac{d}{dt}T(t)w(s-t)-T(t)\frac{d}{dt}w(s-t)\\ &=AT(t)w(s-t)-T(t)Aw(s-t)\\ &=T(t)Aw(s-t)-T(t)Aw(s-t)\\ &=0. \end{align}$$
It follows that $\phi$ is constant, that is, $\phi(t_1)=\phi(t_2)$ for all $t_1,t_2\in[0,s]$. Taking $t_1=0$ and $t_2=s$, we obtain
$$u(s)-v(s)=w(s)=T(0)w(s-0)=\phi(0)=\phi(s)=T(s)w(s-s)=T(s)0=0.$$
Since $s$ is arbitrary, we conclude that $u=v$ on $[0,t_e]$.
Remark: This argument appears in many books in the proof of the following result: If $A:D(A)\subset X\to X$ is the infinitesimal generator of a $C_0$-semigroup $\{T(t)\}_{t\geq 0}$ on $X$, then, for each $U_0\in D(A)$, the abstract Cauchy problem $U_t=AU$, $U(0)=U_0$ has an unique classical solution. Moreover, this solution is given by $U(t)=T(t)U_0$. See, for example, Goldstein's book (page 83) and Zheng's book (page 34). In fact, as we can see in Pazy's book (page 105) , a similar argument can also be applied to prove uniqueness of classical solution for inhomogeneous Cauchy problems.