Is the dominating set problem restricted to planar bipartite graphs of maximum degree 3 NP-complete? Does anyone know about an NP-completeness result for the DOMINATING SET problem in graphs, restricted to the class of planar bipartite graphs of maximum degree 3?

I know it is NP-complete for the class of planar graphs of maximum degree 3 (see the Garey and Johnson book), as well as for bipartite graphs of maximum degree 3 (see M. Chlebík and J. Chlebíková, "Approximation hardness of dominating set problems in bounded degree graphs"), but could not find the combination of the two in the literature.
What if you simply do the following: Given a graph $G = (V,E)$, construct another graph $G' = (V \cup U, E')$ by subdividing each edge of $G$ in 4 parts; here $U$ is the set of new nodes that we introduced, and $|U| = 3|E|$.

The graph $G'$ is bipartite. Moreover, if $G$ is planar and has max. degree 3, then $G'$ is also planar and has max. degree 3.

Let $D'$ be a (minimum) dominating set for $G'$. Consider an edge $(x,y) \in E$ that was subdivided to form a path $(x,a,b,c,y)$ in $G'$. Now clearly at least one of $a,b,c$ is in $D'$. Moreover, if we have more than one of $a,b,c$ in $D'$, we can modify $D'$ so that it remains a valid dominating set and its size does not increase. For example, if we have $a \in D'$ and $c \in D'$, we can equally well remove $c$ from $D'$ and add $y$ to $D'$. Hence w.l.o.g. we have $|D' \cap U| = |E|$.

Then consider $D = D' \cap V$. Assume that $x \in V$ and $x \notin D'$. Then we must have a node $a \in D'$ such that $(x,a) \in E'$. Hence there is an edge $(x,y) \in E$ such that we have a path $(x,a,b,c,y)$ in $G'$. Since $a,b,c \in U$ and $a \in D'$, we have $b, c \notin D'$, and to dominate $c$ we must have $y \in D'$. Hence in $G$ node $y$ is a neighbour of $x$ with $y \in D$. That is, $D$ is a dominating set for $G$.

Conversely, consider a (minimum) dominating set $D$ for $G$. Construct a dominating set $D'$ for $G'$ so that $|D'| = |D| + |E|$ as follows: For an edge $(x,y) \in E$ that was subdivided to form a path $(x,a,b,c,y)$ in $G'$, we add $a$ to $D'$ if $x \notin D$ and $y \in D$; we add $c$ to $D'$ if $x \in D$ and $y \notin D$; and otherwise we add $b$ to $D'$. Now it can be checked that $D'$ is a dominating set for $G'$: By construction, all nodes in $U$ are dominated. Now let $x \in V \setminus D'$. Then there is a $y \in V$ such that $(x,y) \in E$, and hence along the path $(x,a,b,c,y)$ we have $a \in D'$, which dominates $x$.

In summary, if $G$ has a dominating set of size $k$, then $G'$ has a dominating set of size at most $k + |E|$, and if $G'$ has a dominating set of size $k + |E|$, then $G$ has a dominating set of size at most $k$.

Edit: Added an illustration. Top: the original graph $G$; middle: graph $G'$ with a "normalised" dominating set; bottom: graph $G'$ with an arbitrary dominating set.