Similar to the hint in the comment, define $$ L^p (\Bbb {R}^n) \to L^p (\Bbb {R}^n), f \mapsto ( x\mapsto f ((1+\epsilon)x) $$ and show (exercise) that there is $C>0$ with $$ \|T_\epsilon f \|_p \leq C \cdot \|f\|_p \quad \quad (\ast) $$ for all $\epsilon \leq 1$ and all $ f \in L^p (\Bbb {R}^n)$.
Now, I will assume $u \in L^p (\Bbb {R}^n)$ instead of $L_{\rm loc}^p $. The modification for that case is not hard. We have $$ \|\tilde {u_\epsilon} - u\|_p = \|(T_\epsilon u_\epsilon ) -u\|_p \leq \|T_\epsilon (\eta_\epsilon \ast u) -T_\epsilon u\|_p + \| T_\epsilon u - u\|_p \leq C \|u_\epsilon - u\|_p + \|T_\epsilon u - u\| \to 0 $$ as $\epsilon \to 0$.
Here, the last step used $$ T_\epsilon u \to u $$ in $L^p $. This is not hard to see for $u \in C_c $ and follows for general $u $ by approximation (using $(\ast) $ from above).