The answer is: This can be done if and only if the union is convex.

Indeed, let $P:=\{x\colon Ax\le b\}$, $Q:=\{x\colon Bx\le c\}$, and $R:=P\cup Q$. The necessity of the convexity of $R$ was already pointed out by Robert Israel.

Now suppose that $R$ is convex. Note that $P$ and $Q$ are convex polyhedra. Also, any extreme point $p$ of the convex set $R$ is an extreme point of either $P$ or $Q$, depending on whether $p\in P$ or $p\in Q$; similarly for the extreme rays of $R$. So, $R$ is a convex polyhedron and hence the intersection of a finite number of half-spaces; that is, $R=\{x\colon Cx\le d\}$ for some appropriate $C$ and $d$.

**Added in response to a comment by the OP:** If $R$ is convex, then the plane of any face of $R$ is the plane of a face of either $P$ or $Q$, and so, the matrix $[C\ d]$ is a row-submatrix of the matrix
$\begin{bmatrix}A&b\\B&c\end{bmatrix}$. Indeed, take any support plane $S$ of the convex polyhedron $R$. Then $S$ is a support plane of both $P$ and $Q$. Suppose now that $F:=S\cap R$ is of dimension $n-1$, where $n$ is the maximum of the dimensions of $P$ and $Q$; that is, $F$ is a face of $R$. Then either $S\cap P$ or $S\cap Q$ is of dimension $n-1$, and hence a face of $P$ or $Q$, as desired.