Theorem 1. The problem admits a 2-approximation algorithm that runs in $O((m+n)\log n)$ time, given a graph $G=(V,E)$ with $m$ edges and $n$ vertices.
[Caveat: The current post doesn't specify the objective-function value if one or both of the clusters contains no edges. I assume that the objective-function value only sums the maximum-weight edges within clusters that do contain edges. (If, say, the graph is bipartite then the optimal value is zero.)]
Proof. Here's the algorithm:
let $e_1, e_2, \ldots, e_m$ denote the edges sorted by decreasing weight
let $G_t=(V, E_t)$ where $E_t=\{e_1,e_2,\ldots, e_t\}$
denote the graph with only the heaviest $t$ edges
let $t'\in\{1,\ldots, m\}$ be maximum such that $G_{t'}$ is bipartite
(find $t$ using binary search)
let $(C_1, C_2)$ be a bipartition of $G_{t'}$ (such that $E_{t'}\subseteq C_1\times C_2)$)
return $(C_1, C_2)$
The algorithm can be implemented to run in $O((m+n)\log n)$ time, because the binary search requires $O(\log m) = O(\log n)$ rounds, and each round requires checking whether a given $G_t$ is bipartite (and finding its bipartition, if it is), which can be done in $O(n+m)$ time using, say, depth-first search (see e.g. here).
Consider any execution of the algorithm. Let $t'$ and $(C_1, C_2)$ be as computed by the algorithm. To finish we show that $(C_1, C_2)$ has objective-function value at most twice the optimum.
If the given graph $G$ is bipartite, then the algorithm returns a bipartition of $G$, which (by the caveat above) is an optimal solution. So assume that $G$ is not bipartite. So $t' < m$.
Each edge within $C_1$ or $C_2$ is not in $E_{t'}$, so has weight at most $w(e_{t'+1})$. So the algorithm's solution achieves objective-function value at most $2 w(e_{t'+1})$.
Now consider any optimal solution $(C^*_1, C^*_2)$. Because $G$ is not bipartite, there is at least one edge within one of the clusters $C^*_1$ or $C^*_2$. Let $e_{t^*}$ be the edge with maximum minimum index (and hence maximum weight) within either cluster. The value of the optimal solution is at least $w(e_{t^*})$ (using here that the edge weights are non-negative).
Removing $e_{t^*}$ and all cheaper edges yields the graph $G_{t^*-1}=(V, E_{t^*-1})$. By the choice of $e_{t^*}$ this graph is bipartite with bipartition $(C^*_1, C^*_2)$ (as all edges within each cluster are not in $E_{t^*-1}$). Hence $t^*-1 \le t'$, and $w(e_{t^*}) \ge w(e_{t'+1})$. Hence the optimal solution value is at least $w(e_{t'+1})$.
$~~~~~\Box$
