The following is an improved version of a proof I gave in a question I asked on Math Overflow about some generalizations of the present question.
Theorem. A graph (finite or infinite) without isolated vertices has a minimal edge cover.
Proof. Let $G=(V,E)$ be a graph without isolated vertices. Let $M$ be a maximal matching
in $G$, and let $W$ be the set of vertices not covered by $M$. For each vertex $w\in W$, choose an edge $f_w\in E$ which is incident with $w$. Let $S=\{f_w:w\in W\}$ and let $M'$ be the set of all edges $e\in M$ such that at least one endpoint of $e$ is not covered by $S$. Then $S\cup M'$ is a minimal edge cover of $G$.
A set of vertices whose (closed) neighborhoods partition $V(G)$ is called a perfect code (sometimes also an efficient dominating set). See for example Chapter 4 in this book.
The term was first introduced here, extending the notion of perfect codes from coding theory. In general a perfect $d$-code in a metric space is a covering with disjoint balls of radius $d$. In graphs the metric is just the usual distance, so balls of radius 1 are the same as neighborhoods $N[v]$ of vertices. In coding theory the vertices would correspond to words and the distance is for example Hamming distance; the idea is that if you select a vertex in the perfect code, then even after making one step in error, you will always be able to recover it (in a unique way).
Perfect codes are always independent sets (no two elements are adjacent) and dominating sets (their neighborhoods cover everything). Deciding whether any perfect code exists in a given graph is NP-hard, even in planar graphs (Theorem 4.4 in the book). If there is one, then all perfect codes have the same size (Theorem 4.2).
However, note that in the Lights Out puzzle you cite you don't actually need a perfect code. You need a set of vertices $S$ such that every vertex is included in $N[v]$ for $v \in S$ an odd number of times (instead of exactly once). The condition can be phrased as a system of linear equations over $\mathbb{Z}_2$, which can be solved in polynomial time! See for example here.
Best Answer
This set is called a dominating set and the respective algorithmic problem is NP-complete, see, for instance, Wikipedia.