If $\mathcal{O}$ denote the collection of real-valued functions defined on $[0,1]$ that map every set of measure zero to a set of measure 0.
can I describe the set $\mathcal O$ by the following statement? $$m^*(f(E)) < \epsilon \quad \text{if} \quad m(E) < \delta $$
Let $m^*f$ be the first measure on $[0,1]$, such that $m^*f(E)=m^*(f(E))$ and $m$ the second measure. Then we assume that $m^*f \ll m$ (i.e. that $m^*f$ is absolutely continuous with respect to $m$).
If $m^*f([0,1])<\infty$, then the Radon-Nikodym theorem says that $m^*f \ll m$ is equivalent to the following stronger statement: for any $\varepsilon > 0$ there is a $\delta > 0$ such that $m^*f(E)<\varepsilon$ for every $E$ with $m(E)<\delta$
See Link for source.
So yes, you can describe $\mathcal{O}$ like you say if you assume that $m^*f$ is finite. I am not sure, but maybe it is enough if it is $\sigma$-finite.