Notation. For a graph $(V,E)$, we call $e=|E|, v=|V|$.
In a planar graph, we have a maximum number of edges: $e\leq 3v-6$ if $v\geq 3$. Thus, for a graph of thickness $2$, we have the trivial upper bound $$ e\leq 6v-12. $$
Here I will prove the following.
Proposition. Let $(V,E)$ be a graph with thickness $2$ and such that $e\in\{6v-14,6v-13,6v-12\}$. Then for any decomposition $E=E_1\cup E_2$ such that the graphs $(V,E_1),(V,E_2)$ are planar, the two are also connected.
That is, any graph with thickness two such that $e=6v-14$ (or higher) will be an answer to your problem.
Proof. To see that, assume you decompose $E=E_1\cup E_2$. > Since $(V,E_i)$ must be planar graphs (and $v\geq 5$), we have $$|E_i|\leq 3|V|-6$$ Since $|E|=6|V|-14$ and $|E_1|+|E_2|=|E|$, it follows that $$ |E_i|\geq 3|V|-8. $$ Now, I claim that the two subgraphs cannot be disconnected. > In fact, to any disconnected planar graph with $v\geq 4$ you can add at least three edges while still keeping it planar (the main reason is that any face of a planar graph has at least three vertices). But if you add three edges to $E_i$ you make it exceed the maximum number of edges for a planar graph, so both $(V,E_1)$ and $(V,E_2)$ must be connected.
The question would then be: is there a graph with thickness two and with $e=6v-14$? For that to be true, you need at least $|V|\geq 11$ (as $e\leq {v\choose 2}$, and $6v-14\leq {v\choose 2} \,\iff\,v\geq 11$ if $v\geq 5$). Honestly I have no idea, so it could be trivially impossible.