One of the reasons relativistic theories are so restrictive is because of the rigidity of the the symmetry group. Indeed, the (homogeneous part) of the same is simple, as opposed to that of non-relativistic systems, which is not.
The isometry group of Minkowski spacetime is
\begin{equation}
\mathrm{Poincar\acute{e}}=\mathrm{ISO}(\mathbb R^{1,d-1})=\mathrm O(1,d-1)\ltimes\mathbb R^d
\end{equation}
whose homogeneous part is $\mathrm O(1,d-1)$, the so-called Lorentz Group1. This group is simple.
On the other hand, the isometry group of Galilean space+time is2
\begin{equation}
\text{Bargmann}=\mathrm{ISO}(\mathbb R^1\times\mathbb R^{d-1})\times\mathrm U(1)=(\mathrm O(d-1)\ltimes\mathbb R^{d-1})\ltimes(\mathrm U(1)\times\mathbb R^1\times\mathbb R^{d-1})
\end{equation}
whose homogeneous part is $\mathrm O(d-1)\ltimes\mathbb R^{d-1}$, the so-called (homogeneous) Galilei Group. This group is not semi-simple (it contains a non-trivial normal subgroup, that of boosts).
There is in fact a classification of all physically admissible kinematical symmetry groups (due to Lévy-Leblond), which pretty much singles out Poincaré as the only group with the above properties. There is a single family of such groups, which contains two parameters: the AdS radius $\ell$ and the speed of light $c$ (and all the rotation invariant İnönü-Wigner contractions thereof). As long as $\ell$ is finite, the group is simple. If you take $\ell\to\infty$ you get Poincaré which has a non-trivial normal subgroup, the group of translations (and if you quotient out this group, you get a simple group, Lorentz). If you also take $c\to\infty$ you get Bargmann (or Galilei), which also has a non-trivial normal subgroup (and if you quotient out this group, you do not get a simple group; rather, you get Galilei, which has a non-trivial normal subgroup, that of boosts).
Another reason is that the postulate of causality is trivial in non-relativistic systems (because there is an absolute notion of time), but it imposes strong restrictions on relativistic systems (because there is no absolute notion of time). This postulate is translated into the quantum theory through the axiom of locality,
$$
[\phi(x),\phi(y)]=0\quad\forall x,y\quad \text{s.t.}\quad (x-y)^2<0
$$
where $[\cdot,\cdot]$ denotes a supercommutator. In other words, any two operators whose support are casually disconnected must (super)commute. In non-relativistic systems this axiom is vacuous because all spacetime intervals are timelike, $(x-y)^2>0$, that is, all spacetime points are casually connected. In relativistic systems, this axiom is very strong.
These two remarks can be applied to the theorems you quote:
Reeh-Schlieder depends on the locality axiom, so it is no surprise it no longer applies to non-relativistic systems.
Coleman-Mandula (see here for a proof). The rotation group is compact and therefore it admits finite-dimensional unitary representations. On the other hand, the Lorentz group is non-compact and therefore the only finite-dimensional unitary representation is the trivial one. Note that this is used in the step 4 in the proof above; it is here where the proof breaks down.
Haag also applies to non-relativistic systems, so it is not a good example of OP's point. See this PSE post for more details.
Weinberg-Witten. To begin with, this theorem is about massless particles, so it is not clear what such particles even mean in non-relativistic systems. From the point of view of irreducible representations they may be meaningful, at least in principle. But they need not correspond to helicity representations (precisely because the little group of the reference momentum is not simple). Therefore, the theorem breaks down (as it depends crucially on helicity representations).
Spin-statistics. As in Reeh-Schlieder, in non-relativistic systems the locality axiom is vacuous, so it implies no restriction on operators.
CPT. Idem.
Coleman-Gross. I'm not familiar with this result so I cannot comment. I don't even know whether it is violated in non-relativistic systems.
1: More generally, the indefinite orthogonal (or pseudo-orthogonal) group $\mathrm O(p,q)$ is defined as the set of $(p+q)$-dimensional matrices, with real coefficients, that leave invariant the metric with signature $(p,q)$:
$$
\mathrm O(p,q):=\{M\in \mathrm{M}_{p+q}(\mathbb R)\ \mid\ M\eta M^T\equiv \eta\},\qquad \eta:=\mathrm{diag}(\overbrace{-1,\dots,-1}^p,\overbrace{+1,\dots,+1}^q)
$$
The special indefinite orthogonal group $\mathrm{SO}(p,q)$ is the subset of $\mathrm O(p,q)$ with unit determinant. If $pq\neq0$, the group $\mathrm{SO}(p,q)$ has two disconnected components. In this answer, "Lorentz group" may refer to the orthogonal group with signature $(1,d-1)$; to its $\det(M)\equiv+1$ component; or to its orthochronus subgroup $M^0{}_0\ge+1$. Only the latter is simply-connected. The topology of the group is mostly irrelevant for this answer, so we shall make no distinction between the three different possible notions of "Lorentz group".
2: One can prove that the inhomogeneous Galilei algebra, and unlike the Poincaré algebra, has a non-trivial second co-homology group. In other words, it admits a non-trivial central extension. The Bargmann group is defined precisely as the centrally extended inhomogeneous Galilei group. Strictly speaking, all we know is that the central extension has the algebra $\mathbb R$; at the group level, it could lead to a factor of $\mathrm U(1)$ as above, or to a factor of $\mathbb R$. In quantum mechanics the first option is more natural, because we may identify this phase with the $\mathrm U(1)$ symmetry of the Schrödinger equation (which has a larger symmetry group, the so-called Schrödinger group). Again, the details of the topology of the group are mostly irrelevant for this answer.
Best Answer
If you consider the Dirac field $\psi$ as a linear combination of the particle annihilation part $\psi^+$ and the antiparticle creation part $\psi^{-c}$, then we have the requirement that $P\psi(x)P^{-1}$ should be proportional to $\psi(Px)$, where $P$ is the parity/space inversion operator. (I commit a slight abuse of notation here - the $P$ is both the unitary operator representing parity acting on the space of states and the space inversion operator on Minkowski space sending $(t,\vec x)$ to $(t,-\vec x)$ in the argument.
In general, we have also that this should hold for $\psi^+$ and $\psi^{-c}$, i.e. \begin{align} P \psi^+(x)P^{-1} & = \eta^\ast b_u \psi^+(Px) \\ P \psi^{-c}(x)P^{-1} & = \eta^c b_v \psi^{-c}(Px) \end{align} where the $\eta^\ast$ and $\eta^c$ are the phases by which the anti-creation /annihilation operators transform and the $b_{u/v}$ are the phases by which the fundamental spinors traditionally denoted as $u_s,v_s$ transform. These formulae come about because e.g. $\psi^+(x) = \sum_s \int u_s(\vec p) \exp(-\mathrm{i}px)a_s(\vec p)\mathrm{d}^3p$ schematically, and likewise for $\psi^{-c}$. One can then show that $b_u = -b_v$, and for $P\psi(x)P^{-1}$ to be propertional to $\psi(Px)$ we must then have that $\eta^\ast = -\eta^c$ because otherwise $\psi(x) = c_1 \psi^+(x) + c_2 \psi^{-c}$ cannot be proportional because the relative sign would change under parity.
For a more detailed derivation, see "The quantum theory of fields" by Weinberg, volume 1, section 5.5 as suggested in a comment by TwoBs.