As the other answers note, you're perfectly free to consider (or exclude from consideration) weighted graphs with zero-weight edges.
That said, in my experience, the usual convention in most applications of weighted graphs is to make no distinction between a zero-weight edge and the absence of an edge. One reason for this is that, typically, weighted graphs show up as generalizations of multigraphs, which in turn are generalizations of simple graphs.
Specifically, a multigraph is a graph that (unlike a simple graph) allows multiple edges between the same pair of nodes. Whereas, in a simple graph, any pair of nodes is always connected by 0 or 1 edges, a pair of nodes in a multigraph may be connected by 0, 1, 2, 3 or more (but always a non-negative integer number of) edges.
Generalizing a multigraph to allow for a fractional number of edges between a pair of nodes then naturally leads one to consider weighted graphs, and many algorithms that work on arbitrary multigraphs can also be made to work on such weighted graphs. But for such algorithms, the "weight" of an edge really denotes its multiplicity. Thus, given this interpretation, there can be no meaningful distinction between "no edge" and "0 edges" between a pair of nodes: both mean exactly the same thing.
Of course, a "weighted graph" by definition is really just a graph with a number associated to each edge, and it's perfectly possible to interpret the weight as something other than multiplicity, in which case a distinction between no edge and a zero-weight edge may indeed be meaningful. But trying to apply standard multigraph algorithms to such "strangely weighted graphs" is unlikely to produce results that would make sense in terms of the alternative (non-multiplicity) interpretation of edge weights.