Kind of. But we're going to use the usual computer sciency way of describing this, using the language of binary relations.
You're probably already familiar with binary relations, like equality $=$, less-than-or-equal-to $\le$, subset $\subseteq$, and so on. In general, a binary relation $R$ over a set $X$ is a subset $R \subseteq X \times X$. If $(x,y) \in R$, we denote this as $xRy$.
If $\forall x \in X, xRx$, then $R$ is reflexive. The relations $=$ and $\le$ are reflexive, but $\lt$ is not.
If $\forall x, y, z \in X, xRy\,\wedge\,yRz \Rightarrow xRz$, then $R$ is transitive. Plenty of relations are transitive, including all of the ones given above. If $x \le y$ and $y \le z$, then $x \le z$.
Given a relation $R$, the reflexive transitive closure of $R$, denoted $R^*$, is the smallest relation $R^*$ such that $R \subseteq R^*$, and $R^*$ is reflexive and transitive.
Interpreting your graph as a binary relation (since the edges don't really seem to matter to you, you're only interested in the set of vertices), this is exactly what you want: $xR^*y$ if and only if $y$ is a "descendant" (by your meaning) of $x$.
When looking at the literature, you will need to know one more piece of notation: the transitive closure of $R$, denoted $R^+$, is the smallest relation $R^+$ such that $R \subseteq R^+$, and $R^+$ is transitive. Algorithms for computing the transitive closure and the reflexive transitive closure are related, since they differ only by the "diagonal" entries: $R^+ \cup \left\{ (x,x)\,|\,x \in X \right\} = R^*$.
There are several standard algorithms for computing the RTC of a relation. If the relation is dense, in the sense that it's feasible to represent it as a bit matrix, the Floyd-Warshall algorithm is one of the fastest practical algorithms; its run time is $\Theta(|V|^3)$ in theory, but the inner loop is quite fast on real hardware given that it it a bunch of bit vector manipulations.
For sparse relations, see Esko Nuutila's thesis, which contains a very good survey as well as some more recent algorithms.