This is a really hard problem for general graphs.
To give a sense, let's look at the behavior of $E(G)$, the number of edge coverings of $G$, on a couple of simple families of graphs.
First consider $A_n$, the connected graph on $n$ vertices in which one vertex has degree $n-1$ and all other vertices have degree $1$.
Of course we have
$$E(A_n) = 1$$
Next consider the linear graphs $L_n$, which are the cyclic graphs $C_n$ with one edge removed.
For this family one can show via simple inductive argument that the number of edge coverings is $$E(L_n) = F_{n-1} \sim \frac{\varphi^{n-1}}{\sqrt{5}}$$ where $F_n$ is the $n$th Fibonacci number and $\varphi$ is the golden ratio. (Similarly $E(C_n)$ is the $n$th Lucas number.)
Lastly consider the complete graphs $K_n$, for which one can show that the number of edge coverings are $$E(K_n) = \sum\limits_{j=0}^{n}(-1)^j{n \choose j} 2^{n-j \choose 2} \sim 2^{\frac{n(n-1)}{2}}$$
Note $A_n$ and $L_n$ have the same number of vertices and edges but totally different numbers of edge coverings, showing that we need detailed information about at least the degree sequence of a graph. A degree $0$ (isolated) vertex forces $E(G) = 0$ immediately; a degree $1$ vertex contributes no freedom to the edge covering; but increasing a degree $1$ vertex to degree $2$ more or less doubles the number of edge covers, i.e. the problem is inherently exponential.
$E(K_n)$ gives an upperbound on $E(G)$ for $|G| = n$, if we are considering graphs in which the edge set is not allowed to be a multiset.
You might want to check out this paper: https://cs.uwaterloo.ca/journals/JIS/VOL11/Tauraso/tauraso18.html
