Ugh I just lost my post but the short version is that on top of Igor's answer, it is easy to prove this using Edmonds' characterization of the perfect matching polytope, which implies putting weight 1/k on every edge will give you a vector in the polytope. From this fact the matching-coveredness is straightforward.

EDIT:

Edmonds proved that a vector (i.e. an edge-weighting $w(e)$) is in the perfect matching polytope (i.e. the convex hull of incidence vectors of perfect matchings) if and only if the following hold:

1) Every edge has weight in $[0,1]$.

2) Every set $S$ of vertices with odd size has $\sum_{e\in\delta(S)} w(e) \geq 1$, where $\delta(S)$ is the set of edges with exactly one endpoint in $S$.

3) Every vertex $v$ satisfies $\sum_{e\in\delta(\{v\})} w(e) = 1$.

It is an easy exercise show that these conditions are necessary, but as Edmonds proved, they are also sufficient. This implies immediately that if $G$ is a $k-1$-edge-connected graph that is $k$-regular, the vector with every edge getting weight $1/k$ is in the perfect matching polytope of $G$ (in other words, $G$ is fractionally $k$-edge-colourable). Since the weight vector is nonzero everywhere, every edge must be contained in at least one perfect matching. (Again in other words, since only perfect matchings can be used to fractionally $k$-edge-colour a $k$-regular graph, every edge must be in a perfect matching.)