Although you do not seem to require that the triangulation of the closure to
be compatible with the triangulation of the boundary, it is
true in $\mathbb{R}^3$ that a triangulated polyhedron $P$ has a compatible
interior tetrahedralization. Bern proved that, if $P$ has $n$ vertices,
such a tetrahedralization by $O(n^2)$ tetrahedra exists (and can be found quickly):

Bern, Marshall. "Compatible tetrahedralizations." *Fundamenta Informaticae* 22.4 (1995): 371-384. (ACM link.)

In fact, he proved all of $\mathbb{R}^3$ can be tetrahedralized compatible with
$P$'s surface triangulation (with some tetrahedra having a vertex at $\infty$).

It is interesting that if you change "triangulable" to "hexahedral-able,"
and ask if the surface mesh can be extended compatibly to an interior mesh,
the answer is unknown:

"No algorithm is known to construct
hexahedral meshes compatible with an arbitrary given quadrilateral
mesh, or even to determine when a compatible hex mesh
exists, even for the simple examples shown in Figure 1"

Erickson, Jeff. "Efficiently hex-meshing things with topology." *Discrete & Computational Geometry* 52.3 (2014): 427-449. (PDF download.)