It sounds like you may be interested in hypergraphs which is basically what you are asking about except edges can have any number of vertices associated with them. It turns out though that you can model hypergraphs with bipartite graphs. Say you have a bipartite graph with vertices coloured white and black, then a hypergraph can be associated with that by saying at the white vertices all the edges connected to that make the "hyperedge" and the black vertices would be the vertices of the hypergraph. If you have a hypergraph you can get a bipartite graph by adding the white vertices and splitting up the hyperedges into normal edges.
Let $G$ be a $d$-regular bipartite graph. And let $e_{uv}$ be an edge of $G$.
Let $S$ the set of $e_{uv}$ and all its adjacent edges. Then the size of $S$ is $2d-1$ : They are $d$ edges connected at the vertex $u$, and $d$ connected at vertex $v$. As we double count $e_{uv}$, we reduce the count by one, and $|S|=2d-1$.
Recall that an antimatching $A$ of $G$ is a set of edges such that two edges are at most at distance 2, and that the distance between two edges is defined as the distance between corresponding vertices in $L(G)$ (the Line graph of $G$).
As all edges from $S$ are connected to $e_{uv}$, then the maximum distance between edges of $S$ is 2.
Therefore $S$ is an antimatching of size $2d-1$.
I think that's all to proove. (I never studied antimatching before, I might be missing something).
Note An antimatching can also be defined (equivalently) as a subgraph with no induced matching of size greater than 1. This is also clearly the case for $S$ :
Suppose there exist an induced matching of size 2 in $S$. Then it must contains one edge connected to $u$ and one connected to $v$. But then in order to be induced we would need to include $e_{uv}$, and would not be a matching. Hence a contradiction.
Therefore the only induced matching of $S$ have size 1.
Best Answer
You don't even need a pair. A graph may be defined simply as a set $G$ of singleton or two-element sets. The set of vertices is then $V=\bigcup G$, while $E$ is the set of two-element members of $G$, and $G\setminus E$ is the set of isolated vertices. However, this will not do if you need to specify loops, which are edges that connect a vertex to itself, or multiple edges, although these types of edge would be regarded as illegitimate or irrelevant in the mainstream of graph theory. And of course ordered pairs are needed to define a directed graph.