I want to prove the following, $$ \lim_{t->0} \|T_p(t)f-f\|_p = 0, $$ where $T_p(t)f(s) = \chi_{[0,e^{-t}]}(s)e^{t/p}f(se^t)$, with $f\in L^p([0,1])$ and $t>0$ and $s \in [0,1]$. In fact, we have that,
$$ \begin{aligned} \lim_{t->0} \|T_p(t)f-f\|_p & = \left( \lim_{t->0} \int_0^{e^{-t}} |e^{t/p} f(se^t) - f(s)|^p \, ds + \lim_{t->0} \int_{e^{-t}}^{1} |f(s)|^p \, ds \right)^{1/p} \\ &= (L_1+L_2)^{1/p}. \end{aligned} $$
On one hand as $f\in L^p([0,1])$ we can use dominated convergence theorem so, $L_2 = 0$. On the other hand $L_1 \leq 2^p\|f\|_p^p$ using Minkowski inequality therefore we can use again the dominated convergence theorem so,
$$ \begin{aligned} L_1 & = \lim_{t->0} \int_0^{e^{-t}} | f(se^t) - f(s)|^p \, ds = \int_0^{1} \lim_{t->0} | f(u) - e^{-t/p}f(e^{-t}u)|^p \, du \\ &= \int_0^{1} | f(u) - \lim_{t->0} f(e^{-t}u)|^p \, du. \end{aligned} $$
But here I face the problem to check that $\lim_{t->0} f(e^{-t}u) = f(u)$. Can I assure that this is true just because $f \in L^p([0,1])$?
Okay I think I got the solution following the hints of @mihaild.
As we know the set of simple functions is dense in $L^p([0,1])$ we assume $f$ to be a simple function such that $f(u) = \sum_{i=0}^n \lambda_k \chi_{I_k}$ where $I_k=[a_k,a_{k+1})$ for $k \in \{1,2,\dots, n-1\}$ and $I_n=[a_n, a_{n+1}]$ with $a_0=0$ and $a_{n+1}=1$.
Therefore,
$$ \lim_{t\to 0} \int_0^1 |f(u)- e^{-t/p}f(e^{-t}u)|^p \, du = \lim_{t\to 0} \sum_{k=0}^n \int_{a_k}^{a_{k+1}} |f(u)- e^{-t/p}f(e^{-t}u)|^p \, du. $$
Now we take $t$ as small as possible in order to assure that $a_{k-1} < e^{-t} a_k$ and that $a_k e^t< a_{k+1}$ for every $k$. Notice that this is possible because we are dealing with a finite amount of intervals. Thus,
$$ \lim_{t\to 0} \int_0^1 |f(u)- e^{-t/p}f(e^{-t}u)|^p \, du = \lim_{t\to 0} \sum_{k=0}^n \int_{a_k}^{a_k e^t } |\lambda_k- e^{-t/p} \lambda_{k-1}|^p \, du + \int_{a_k e^t}^{a_{k+1} } |\lambda_k- e^{-t/p} \lambda_{k}|^p \, du. $$
Where $\lambda_{-1}$ has to be understood as 0. Therefore,
$$ \begin{aligned} \lim_{t\to 0} \int_0^1 |f(u)- e^{-t/p}f(e^{-t}u)|^p \, du & = \lim_{t\to 0} \sum_{k=0}^n a_k (e^t-1) | \lambda_k- e^{-t/p} \lambda_{k-1}|^p \\ & + \lim_{t\to 0} \sum_{k=0}^n (a_{k+1}-a_k e^t) |\lambda_k|^p|1- e^{-t/p}|^p\\ & = 0 \end{aligned} $$
because in the first sum $e^t-1 \to 0$ as $t\to 0$ and in the second sum $1-e^{-t/p} \to 0$ as $t \to 0$.
This completes the proof.