I would call these orders *self-similar*, since the whole order consists of two copies of itself, in the way of many fractals.

Examples would include the rational line
$\langle\mathbb{Q},\lt\rangle$ as well as the Cantor set and many
other fractal-like orders.

You asked,

is it a specific class of ordinal or whatever?

Note that no (nonzero) ordinal has your property, since no
ordinal is isomorphic to a proper initial segment of itself.
Indeed, one may easily iterate the isomorphism of $A$ with $B$ to
produce an infinite descending sequence, so if nonempty, it cannot
be well-ordered.

It is easy to construct such orders, and all such orders arise by
a suitable iteration of the following procedure:

We have the current collection of points in $B$ and $C$, with
every point in $B$ below every point in $C$, and we have
order-preserving maps from $B\sqcup C$ to $B$ and also to $C$.

- We add a new point $p$ either to $B$ or $C$.
- This new point realizes a certain cut in the order we have so far.
- We extend the maps by considering the image and pre-image of that cut, creating further new points if necessary to realize those
cuts, closing under this process.

For example, one can start with one point in $B$, but this causes
one to add a point to $C$ to which to map it, which causes one to
create the image of *that* point in $B$, and so on, back and
forth. One can control this process by ensuring during the
construction that certain cuts will or will not have a least upper
bound.

Finally, note that the property can have no consequences as to the purely local nature of the order, since if we have a self-similar order, we can replace each point by a copy of any fixed linear order, and this resulting order will also be self-similar. For example, $\mathbb{Q}$ copies of $L$, for any linear order $L$, is self-similar. But this order is, locally, just like $L$.