There is nothing to say in the general case, but in the case of a *continuous* dcpo $D$ there is a well-known construction of certain bases for the Scott topology on $D$.

We say $d$ is way below $e$, or $d \ll e$ if for each directed set $(d_i)_{i \in I}$ such that $e \leq \bigvee_{i \in I}d_i$, there exists $i \in I$ such that $d \leq d_i$. For example, if $\mathcal{O}(X)$ is the lattice of open sets of a topological space $X$, then if $U,V$ are open sets, and $K \subseteq X$ is a compact subspace such that $U \subseteq K \subseteq V$, then $U \ll V$ in $\mathcal{O}(X)$.

For each $d \in D$ in a dcpo, define
$$
\mathbf{waydown}(d) = \{ e \in D \mid e \ll d \}.
$$
This is usually written like $\downarrow d$ but with a double arrowhead, but I couldn't get this symbol on MathOverflow.

A dcpo $D$ is said to be *continuous* if for all $d \in D$ the set $\mathbf{waydown}(d)$ is directed and $d = \bigvee\mathbf{waydown}(d)$.

We can now define
$$
\mathbf{wayup}(d) = \{e \in D \mid d \ll e \}
$$
In a continuous dcpo $D$, each $\mathbf{wayup}(d)$ is Scott open, and these sets form a base for the Scott topology on $D$.

Additionally, there is the notion of a *basis for a dcpo*. This is a set $B \subseteq D$ such that for all $d \in D$, $B \cap \mathbf{waydown}(d)$ is directed and has supremum $d$. If $B$ is a basis for $D$, then $\{\mathbf{wayup}(b)\}_{b \in B}$ is a base for the Scott topology of $D$.

You can find more information on this sort of thing in the book *Continuous Lattices and Domains* by Gierz, Hoffman, Keimel, Lawson, Mislove and Scott, particularly section III-4.