$\newcommand\de\delta\newcommand\ol\overline$Your goal cannot be attained in general. Indeed, suppose that $\de\in(0,1/2)$. Take any interval $E$ of length $\de_1:=|E|\in(\de,2\de)$. Then for the closure $\ol E$ of $E$ and some closed intervals $I_1$ and $I_2$ of lengths $|I_1|=\de$ and $|I_2|=\de_1-\de\le\de$ we have $E=I_1\cup I_2$, so that 
$$\mu_\de^s(E)\le|I_1|^s+|I_2|^s
=\de^s+(\de_1-\de)^s\underset{\de_1\downarrow\de}\longrightarrow
\de^s<\de^{s-1}-\de^s,$$
so that for some interval $E$ with $\de<|E|$ we have 
$$\mu_\de^s(E)\not\ge\de^{s-1}-\de^s.$$

(It is actually easy to see that for any interval $E$ of length $t$ we have $\mu_\de^s(E)=k\de^s+(t-k\de)^s$, where $k:=\lfloor t/\de\rfloor$.)