What is the complexity of this weighted b-edge matching problem?
Assuming $x(e)=1$ in Condition 2, the problem is NP-complete. 

Clearly it is in NP.  We show NP-hardness by reduction from Subset Sum:

Lemma 1. The problem is NP-hard.

Proof. The proof is by the following reduction from Subset Sum.  Given a Subset-Sum input $(y, T)$, where $y=(y_1, y_2, \ldots, y_n)$ is a sequence of integers, and $T$ is the target, the reduction outputs the following instance of the problem defined by OP.

For each given integer $y_i$, create two new vertices $v_i$ and $v'_i$, and add two copies of edge $e_i=(v_i, v'_i)$, each with weight $w(e_i) = y_i$.  Put one copy of $e_i$ in $E_1$ and the other in $E_2$, and make $b_\ell(v_i) = b_u(v_i) = b_\ell(v'_i) = b_u(v'_i) = 1$.  Finally, take $r_1 = T$ and $r_2 = (\sum_i y_i) - T$.  This completes the reduction.

Given any subsequence of $y$ that sums to $T$, there is a corresponding solution $(E'_1, E'_2)$ to the created instance, where, for each $i$, if $y_i$ is in the subsequence we put the copy of $e_i$ that is in $E_1$ in $E'_1$, and otherwise we put the copy of $e_i$ that is in $E_2$ in $E'_2$.  Then each vertex has one chosen edge incident to it, the total weight of edges in $E'_1$ is $T$, and the total weight of edges in $E'_2$ is $(\sum_i y_i) - T$, as required.

Conversely, consider any solution $(E'_1, E'_2)$ to the created instance.  Form a corresponding solution to the Subset-Sum instance by choosing the elements $y_i$ of $y$ such that $e_i$ is in $E'_1$.  It sums to at least $T$ by Condition 1. And it sums to at most $T$ because, for each edge $e_i$ that is not in $E'_1$, it must be that $e_i$ is in $E'_2$ (by Condition 2), and the total weight of such edges must be at least $(\sum_i y_i) - T$ (by Condition 1).  $~~\Box$