$\pi$, $\sigma^\pm$ components with no magnetic field?
There's several things going on here, and a good bit of misconceptions. The concepts of a $\sigma^\pm$ or a $\pi$ transition are always relative to a given quantization axis. If there is an external magnetic field, then that is implicitly taken as the quantization axis; if there isn't one then it needs to be explicitly set.
It is also important to note that there is no explicit need for the quantization axis to coincide with the external magnetic field. Doing it helps simplify the calculation both mathematically and conceptually, but ultimately it is nothing more than a choice of basis for a vector space and all the physics is unchanged if you change to some awkward basis.
What does happen, however, if you analyze a situation with an external field in some other axis, is that the at-rest Zeeman effect hamiltonian is no longer diagonal so even in the absence of a radiation field you will get Zeeman-induced "transitions" between your basis states, because they are not eigenstates of the atom-only hamiltonian.
OK, so let's get into some of your misconceptions.
A $\sigma^+$ ($\sigma^-$) photon traveling along the $z$ direction (i.e. parallel to the magnetic field) appears as a left (right) circularly polarized wave. But the same photons traveling in the $x$ or $y$ direction appear as linear.
This is only a vague nod in the direction of being correct. There are no $\sigma$ or $\pi$ photons, there are $\sigma$ and $\pi$ transitions, and you normally tend not to define both at the same time. Once you get into the language, then you can extend the term to say "$\pi$ photon" and mean "photon polarized so that it will effect a $\pi$ transition", but the photon itself has linear or circular polarization.
This is why the statement above is just completely wrong. If you have a circularly polarized photon and you change the axis it's propagating on, to be the "same" photon it still needs to be circularly polarized. Conversely, a $\sigma^\pm$ transition can be caused by a $z$-propagating circularly polarized photon as well as by $x$-propagating $y$-polarized photons. See the difference?
Coming now to the crux of the question,
From this it is clear that when observing in the $x$ direction you will only see linearly polarized photons and when in the $z$ direction only circularly polarized photons.
this is roughly true, but you should note that whenever you can see left- as well as right-circular photons you can also see superpositions of the two, which are generally linear polarizations. Similarly, if you observe from the $x$ axis you can see both $y$-polarized (linear) photons coming from $\sigma$ transitions and $z$-polarized photons coming from $\pi$ transitions, but you can also observe their superpositions, which include circular polarizations.
...or, at least, the argument above would work if you were superposing photons at the same frequency. In the presence of a magnetic field, however, the levels are shifted by different amounts depending on the magnetic quantum number $m$ (if you set the quantization axis along the magnetic field), which means that if you observe along the $x$ axis then the $y$-polarized and $z$-polarized components will be at different frequencies; the same holds for the two circular components observed along the quantization axis. You can still observe superpositions of these components, but being at different frequencies they no longer combine into a single component with different polarization.
This then directly answers your question:
Is the same true when there is no magnetic field present or does something change in the analysis meaning we can see linearly polarized photons and circularly polarized photons in both directions? If so what and if not why not?
If there is no magnetic field, then there is no splitting between the levels, and all the radiation is at the same frequency and it can be combined as you wish. This therefore comes in two flavours:
If you're trying to use an external field to stimulate a transition up from an $S$ to a $P$ state, then any polarization will work, and the state of the polarization will determine which state within the $P$ manifold will get populated.
Circular polarizations will populate $m=\pm1$ states about a quantization axis along the propagation direction; linear polarizations will populate $m=0$ states about a quantization axis along the polarization direction.
You're not, of course, beholden to this analysis: you can put your quantization axis wherever you want, and the analysis of which states do or don't get populated will be equivalent, though it may be harder if you choose an awkward basis.
On the other hand, if you already have an excited $P$ state and you're looking for its emission, then the polarization and propagation direction of the emission will depend on the state, pretty much as described in your book if you start with an $m=0$ or $m=\pm1$ state.
And, again, it's important to note that there are other, funkier states, like superpositions of $m=0$ and $m=\pm1$ states, and those will emit in different directions. Naturally, these are some of the states that come into play if you analyze an arbitrary state in an arbitrary frame.
Either way, in the absence of a natural quantization axis which plays a role in the hamiltonian, the analysis is exactly the same in any frame of reference.