A photon with horizontal polarization crossing a PBS is transmitted, while one with vertical polarization is reflected:
This means that if the input photon is horizontally polarized and is sent in the first input mode, that is,
$$\lvert \text{input} \rangle = \lvert 1,H \rangle_{\text{in}},$$
then the output state is simply
$$\lvert \text{output} \rangle = \lvert 1,H \rangle_{\text{out}},$$
that is, a photon in the first output mode with horizontal polarization.
If on the other hand the input is vertically polarized, than such will be the output, but in the other spatial mode:
$$\lvert \text{input} \rangle = \lvert 1,V \rangle_{\text{in}}
\longrightarrow \,
\lvert \text{output} \rangle = \lvert 2,V \rangle_{\text{out}}.$$
In both cases there is no entanglement neither in the input nor in the output.
If on the other hand you had an input photon with polarization
$\lvert + \rangle \equiv \frac{1}{\sqrt2} (\lvert H \rangle + \lvert V \rangle)$:
$$
\lvert \text{input} \rangle = \lvert 1,+ \rangle,
$$
than you would get as output the state
$$
\lvert \text{output} \rangle =
\frac{1}{\sqrt2}( \lvert 1,H\rangle + \lvert 2,V\rangle ),
$$
which is a superposition of the photon being in the first and second output modes.
This state can be said to be entangled, like in the case you mentioned of the BS.
It is worth noting though that when you have a single particle like in this context one usually speaks of the photon being in a superposition of states, instead that of "entanglement between photon and vacuum", being that of entanglement a concept usually related to multiparticle states (though it can be argued that the concept of entanglement also makes sense with single particles, see for example this paper by von Enk).