Not necessarily. Let $X$ be an uncountable set with the discrete topology, and let $\mathcal{E}$ be the collection of singletons, which is a base for the topology, since every set is a union of singletons.

The Borel sets here are the countable and co-countable sets. If $Y$ is a co-countable subset of $X$, then it is open and in $\text{Bor}(\mathcal{E})$, but it is not the union of countably many singletons.

Not necessarily. Let $X$ be an uncountable set with the discrete topology, and let $\mathcal{E}$ be the collection of singletons, which is a base for the topology, since every set is a union of singletons.

The $\sigma$-algebra generated by the singletons, what you call $\text{Bor}(\mathcal{E})$, is the algebra consisting exactly of the countable and co-countable sets. But notice that if $Y$ is a co-countable subset of $X$, then it is open and in $\text{Bor}(\mathcal{E})$, but it is not the union of countably many singletons. So this is a negative instance of your desired property.