Conditions for bipartite graph to be planar with no edges going around the vertices
Your graphs are exactly the graphs of path-width&nbsp;$1$ or, equivalently, the forests each of whose components is a caterpillar.  Caterpillars have two relevant characterizations: 


they're the trees in which there is a single path containing every vertex of degree more than&nbsp;$1$;
they're the trees in which every vertex has at most two non-leaf neighbours.


Lemma 1. Every caterpillar is in your class.

Proof. Let $G$ be a caterpillar and let $P=x_1\dots x_\ell$ be a longest path containing every vertex of degree&nbsp;$2$ or more.  Note that, by maximality, $d(x_1)=d(x_\ell)=1$. We can produce a drawing of&nbsp;$G$ by first drawing&nbsp;$P$ as a zig-zag and then adding the degree-$1$ vertices adjacent to&nbsp;$x_i$ between $x_{i-1}$ and&nbsp;$x_{i+1}$.&nbsp;$\Box$

Lemma 2. Every graph $G$ in your class is acyclic.

Proof. Suppose $G$ contains the cycle $x_1y_1x_2y_2\dots x_ky_kx_1$ and suppose it has a drawing of the required form. W.l.o.g., $x_2$&nbsp;is above&nbsp;$x_1$. But then we must have $y_2$ above&nbsp;$y_1$ since, otherwise, the lines $x_1y_1$ and&nbsp;$x_2y_2$ would cross. By induction, $x_{i+1}$&nbsp;is above&nbsp;$x_i$ for all $i\in\{1, \dots, k-1\}$ and likewise for the&nbsp;$y$'s. But then any line $y_kx_1$ must either leave the region between the two columns of vertices or cross every other edge in the cycle. This contradicts our assumption that the graph has a proper drawing. $\Box$

Lemma 3. Every connected non-caterpillar is not in your class.

Proof. Let $G$ be a connected graph that is not a caterpillar. If it contains a cycle, it is not in your class by Lemma&nbsp;$2$, so we may assume it is a tree.  If it is not a caterpillar, it must contain a vertex&nbsp;$x$ with distinct neighbours $y_1$, $y_2$ and&nbsp;$y_3$, each of which has degree at least &nbsp;$2$.

Suppose we have a drawing of&nbsp;$G$ with the required properties.  W.l.o.g., $y_2$&nbsp;is above&nbsp;$y_1$ and $y_3$&nbsp;is above&nbsp;$y_2$.  Let $z\neq x$ be a neighbour of&nbsp;$y_2$.  The edge&nbsp;$y_2z$ must cross $xy_1$ or&nbsp;$xy_3$, contradicting our assumption that the graph has a drawing of the required form.&nbsp;$\Box$

Theorem. Your class of graphs is exactly the class of forests each of whose components is a caterpillar.

Proof. Let $G$ be a graph.  Clearly, $G$&nbsp;is in your class if, and only if, every component is: if any component cannot be drawn as required, the whole graph cannot; if every component can be drawn as required, then the whole graph can be drawn by arranging the components one above the other.  The result now follows by Lemmas $1$ and&nbsp;$3$.&nbsp;$\Box$

Corollary. Your class of graphs is the class of graphs that do not have $K_3$ or the subdivision of&nbsp;$K_{1,3}$ as a minor.

Proof. These are the obstructions for path-width&nbsp;$1$.&nbsp;$\Box$

These are essentially the obstructions you found: you need $K_3$ rather than $K_4$ because the latter would admit $K_3$ into the class; the subdivision of $K_{1,3}$ is exactly your second obstruction.