I don't know of a way to reduce your problem to a more well known formulation, but I can prove that your problem is NP-hard:
In particular, consider the decision version in which we ask whether it is possible to make $\max_{j\in J}|\bigcup\{S \in G_j\}|$ less than or equal to some threshold $T$.
I prove below by reduction from clique that this decision problem with $k = 2$ is NP-hard.
First the reduction:
Suppose we are given a graph $G = (V, E)$ with $n$ vertices and $m$ edges and a value $c$. We wish to know whether $G$ has a clique of size $c$. We use $G$ and $c$ to construct an instance of your problem.
Define $t_1 = n + m - {c \choose 2}$ and define $t_2 = {c \choose 2} + c$. Then we set the threshold value $T$ to equal $2t_1 + 2t_2$.
We set the universe of elements $U = \bigcup_{i \in I} S_i$ to equal $$\{a_1, a_2, \ldots, a_{T-t_1}\} \cup \{b_1, b_2, \ldots, b_{T-t_2}\} \cup V \cup E$$.
We create exactly $m + 2$ subsets $S_i$:
- We define $S_1$ to be $\{a_1, a_2, \ldots, a_{T-t_1}\} \cup V$.
- We define $S_2$ to be $\{b_1, b_2, \ldots, b_{T-t_2}\}$
- If $E = \{e_1, e_2, \ldots, e_m\}$, then for $i = 1, \ldots, m$ we define $S_{i+2}$ to be $\{e_i, x, y\}$, where $x$ and $y$ are the two endpoints of $e_i$
Now I'll prove correctness:
There are only two $G_j$s. Note that $S_1$ and $S_2$ cannot be in the same $G_j$ because $|S_1 \cup S_2| = 2T -t_1-t_2+n = T + t_1 + t_2 + n > T$. As a result, the two $G_j$s will contain one of $S_1, S_2$ each. Without loss of generality, suppose $G_1$ contains $S_1$ and $G_2$ contains $S_2$.
Let $E_1$ be the set of edges $e_i$ such that $S_{i+2}$ is in $G_1$ and similarly let $E_2$ be the set of edges $e_i$ such that $S_{i+2}$ is in $G_2$.
Then consider the value $|\bigcup\{S \in G_1\}|$. $G_1$ consists of $S_1$ together with the $S_{i+2}$s for those $i$ for which $e_i \in E_1$. Each $S_{i+2}$ has exactly three elements: two vertices and one edge. But $S_1$ already contains all of the vertices. In other words, we see that the set $\bigcup\{S \in G_1\}$ is equal to $\{a_1, a_2, \ldots, a_{T-t_1}\} \cup V \cup E_1$. Then the size of this set is $(T - t_1) + n + |E_1| = T - (n + m - {c \choose 2}) + n |E_1| = T - (m - {c \choose 2}) + |E_1|$.
Next consider the value $|\bigcup\{S \in G_2\}|$. $G_2$ consists of $S_2$ together with the $S_{i+2}$s for those $i$ for which $e_i \in E_2$. Define $V_2$ to be the set of endpoints of edges in $E_2$. Then the set $\bigcup\{S \in G_2\}$ is equal to $\{b_1, b_2, \ldots, b_{T-t_2}\} \cup E_2 \cup V_2$. The size of this set is $(T -t_2) + |E_2| + |V_2| = (T - {c \choose 2} - c) + |E_2| + |V_2|$.
A partition of edges into $E_1$ and $E_2$ solves this problem if and only if both $T - (m - {c \choose 2}) + |E_1|$ and $(T - {c \choose 2} - c) + |E_2| + |V_2|$ are at most $T$. But $$T - (m - {c \choose 2}) + |E_1| \le T$$ is equivalent to $|E_1| \le (m - {c \choose 2})$, which in turn is equivalent to $|E_2| \ge {c \choose 2}$. Furthermore, $$(T - {c \choose 2} - c) + |E_2| + |V_2| \le T$$ is equivalent to $|E_2| + |V_2| \le c + {c \choose 2}$.
In other words, this problem is solvable if and only if we can choose at least ${c \choose 2}$ edges such that the number of chosen edges plus the total number of vertices incident on those edges is at most $c + {c \choose 2}$. It is possible to choose at least ${c \choose 2}$ edges satisfying this constraint if and only if it is possible to choose exactly ${c \choose 2}$ edges satisfying this constraint (since choosing extra edges can only make it harder to satisfy the constraint). Thus, this problem is solvable if and only if we can choose exactly ${c \choose 2}$ edges such that the number of chosen edges plus the total number of vertices incident on those edges is $c + {c \choose 2}$. In other words, this problem is solvable if and only if we can choose exactly ${c \choose 2}$ edges such that the total number of vertices incident on those edges is at most $c$. This is equivalent to saying that the graph $G$ must have a clique of size $c$. Thus the given reduction is answer preserving.
It is also easy to generalize the above reduction to also show the $k > 2$ case hard. For example, to show $k = 3$ hard, we simply add another set $\{c_1, c_2, \ldots, c_T\}$. Then one of the $G_j$s will be forced to contain this $S_i$ and no other, reducing the rest of the problem to the same situation as above.