I just post some trivial remarks. Let me 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 bound
$$ e\leq 2(3v-6). $$

**Here I show that any graph with thickness two such that $e=6v-14$ will be an answer to your problem**. 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.
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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 $6v-14\leq {v\choose 2} \,\iff\,v\geq 11$). Honestly I have no idea, so it could be trivially impossible.