FWIW, your problem is hard to approximate within a multiplicative factor of $n^{1-\epsilon}$ for any $\epsilon>0$.
We show that below by giving an approximation-preserving reduction from Independent Set, for which the hardness of approximation is known.
Reduction from Independent Set
Let undirected graph $G=(V,E)$ be an instance of Independent Set.
Let $d_v$ denote the degree of vertex $v$ in $G$.
Let $n$ be the number of vertices in $G$.
Construct edge-weighted graph $G'=(V',E')$ from $G$ as follows.
Give each edge in $E$ weight 1.
For each non-isolated vertex $v\in V$,
add $d_v-1$ new edges, each with weight $-1$,
to $d_v-1$ new vertices.
For each isolated vertex $v\in V$,
add one new edge of weight 1 to a new vertex.
(Note: each new vertex (in $G'$ but not $G$) has exactly one neighbor,
which is in $G$.)
Lemma. $G$ has an independent set of size $k$ iff
$G'$ (as an instance of your problem) has a solution of value at least $k$.
Proof. Let $S$ be any independent set in $G$.
Then, since the vertices in $S$ are independent in $G'$,
the value of $S$ in $G'$ (by your objective)
is
$$\sum_{v\in S} d_v - (d_v-1) ~=~ |S|.$$
Conversely, let $S$ be a solution for $G'$ of value at least $k$.
Without loss of generality, assume $S$ contains no new vertices.
(Each new vertex $v'$ is on a single edge $(v',v)$.
If $v$ was not isolated in $G$, then the weight of the edge
is $-1$, so removing $v'$ from $S$ increases the value of $S$.
If $v$ was isolated, then the weight of the edge is 1,
so removing $v'$ from $S$ and adding $v$ maintains the value of $S$.)
Without loss of generality, assume that $S$ is an independent set in $G$.
(Otherwise, let $(u,v)$ be an edge such that $u$ and $v$ are in $S$.
The total weight of $v$'s incident
edges in $G'$ is $d_v - (d_v-1) = 1$, so the total weight of $v$'s incident
edges other than $(u,v)$ is at most zero.
Thus, removing $v$ from $S$ would not increase the value of $S$.)
Now, by the same calculation as at the start of the proof,
the value of $S$ is $|S|$. It follows that $|S| \ge k$. QED
As an aside, you might ask instead for an additive approximation,
of, say, $O(n)$ or $\epsilon m$.
It seems possible to me that for your problem even deciding whether
there is a positive-value solution could be NP-hard.