Unfortunately you can't just define the Poincare group to be that, because in the standard treatment it is defined a little bit differently. What you defined is actually the Lorentz group. The Poincare group contains translations as well.
The Lorentz group $O(1,3)$ is the group of all $\Lambda \in GL(4, \mathbb{R})$ such that
$$\Lambda^T \eta \Lambda = \eta,$$
with $\eta = \operatorname{diag}(-1,1,1,1)$. This can be seen as the group of all "changes of orthonormal frames in spacetime".
Remember that one orthonormal reference frame is a set of vectors $\{e_\mu\}$ such that $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. In that sense, given two such frames, the change of frame that takes components in one of them to components in the other is given by these elements.
With this, the proper Lorentz group is $SO(1,3)$ which basically means that you pick all elements of $O(1,3)$ with determinant $+1$. The orthocronous part just means that if $\Lambda \in O(1,3)$ has $\Lambda^0_0 > 0$ then it preserves the sense of time-like vectors.
Some authors seems to include "by default" the orthocronous requirement in the group $SO(1,3)$ (see for example Analysis, Manifolds and Physics by Choquet-Bruhat, vol. 1, page 290). Others leave it separately, so that you end up with a group $SO^+(1,3)$, but this is a question of convention.
Now the Poincare group $P(1,3)$ (which I don't know any standard notation for) is the group of all Lorentz transformations together with all spacetime translations. In other words, we have:
$$P(1,3)= \{(a,\Lambda) : a\in \mathbb{R}^4, \Lambda \in SO(1,3)\}$$
together with the multiplication defined by
$$(a_1,\Lambda_1)\cdot (a_2,\Lambda_2)=(a_1+\Lambda_1 a_2, \Lambda_1\Lambda_2)$$
Think like that: while elements of $SO(1,3)$ relates orthonormal frames with coincident origins, elements of $P(1,3)$ also allows for the shift of origin as well.
The action of $SO(1,3)$ in the Minkowski vector space $\mathbb{R}^{1,3}$ (not to be confused with flat spacetime - this is actually the "model" for spacetime's tangent spaces, which just happens to be possible to identify with spacetime itself in the flat case) is given by usual matrix multiplication, i.e., given $ \Lambda \in SO(1,3)$ you have:
$$\Lambda \cdot v = \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
On the other hand, the action of $P(1,3)$ on the Minkowski vector space $\mathbb{R}^{1,3}$ is characterized by the fact that given $(a,\Lambda)\in P(1,3)$ you have:
$$(a,\Lambda)\cdot v = a + \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
I don't know if it helps, but people like to compare this to the case in $\mathbb{R}^3$ where you have the rotation group $SO(3)$ and the Euclidean group $E(3)$ comprising rotations in $SO(3)$ with translations in $\mathbb{R}^3$ thus forming the group of rigid motions. This could be seen as the analogous construction in spacetime.
EDIT: Regarding the semi-direct product construction mentioned in comments, recall that given groups $N,H$ with $\varphi : H\to \operatorname{Aut}(N)$ a homomorphism into the group of automorphisms of $N$, we can build the semi-direct product as the set $N\times H$ with the product:
$$(a,b)\cdot (c,d)=(a\varphi(b)(c), bd)$$
the resulting group is denoted $N\rtimes H$. In the particular case it is clear that we have this construction with $N = \mathbb{R}^{1,3}$, $H = SO(1,3)$ and $\varphi : SO(1,3)\to \operatorname{Aut}(\mathbb{R}^{1,3})$ given by $\varphi(\Lambda)(v) = \Lambda v$. Thus
$$P(1,3)=\mathbb{R}^{1,3}\rtimes SO(1,3)$$
You dramatically misunderstood and misquoted Ernie's formula (4.17)!
The whole point of his formula is the term you skipped, $𝑂(𝜃^2)$
on the right-hand side,
$$e^{-i\theta\vec{J}\cdot\hat{n}}=\exp{(-i\theta\ n_x J_x)}\exp{(-i\theta\ n_y J_y)}\exp{(-i\theta n_z J_z)} +𝑂(𝜃^2),
$$
which is to say his rotations only compose like this for infinitesimally small rotations, a point he stresses in (4.28)! This is the essence of Lie groups, namely that the non commutativity first emerges at $𝑂(𝜃^2)$.
You may see this by two sequential applications of the CBH composition of exponentials of noncommuting operators, (4.25), namely
$$
\exp{(-i\theta\ n_x J_x)}\exp{(-i\theta\ n_y J_y)}\exp{(-i\theta n_z J_z)} \\ = \exp \bigl( -i\theta ~\hat n\cdot \vec J -i (\theta^2 /2)(n_xn_y J_z- n_x n_zJ_y + n_y n_z J_x )+ O(\theta^3)\bigr) \\ =
\exp(-i\theta ~\hat n\cdot \vec J ) + O(\theta^2).
$$
cf. (4.29).
As far as non commutativity goes, he threw out the baby with the bathwater, but he is illustrating mappings of sums of matrices, not their noncommutative multiplication...
Best Answer
From special relativity we know that a Lorentz transformation: \begin{equation} x'^\mu = \Lambda^\mu {}_\nu x^\nu \end{equation} preserves the distance: \begin{equation} g^{\mu \nu} \Delta x_\mu \Delta x_\nu = g^{\mu \nu} \Delta x_\mu' \Delta x_\nu' \end{equation} The above two equations imply: \begin{equation} g^{\mu \nu} = g^{\rho \sigma}\Lambda_\rho {}^\mu \Lambda_\sigma {}^\nu \end{equation} Now, let us consider an infinitesimal transformation: \begin{equation} \Lambda_\nu {}^\mu = \delta_\nu{}^\mu + \omega_\nu{}^\mu + O(\omega^2) \end{equation} such that we can write: \begin{equation} \begin{aligned} g^{\mu \nu} & = g^{\rho \sigma}\Lambda_\rho {}^\mu \Lambda_\sigma {}^\nu \\& = g^{\rho \sigma} \left( \delta_\rho{}^\mu + \omega_\rho{}^\mu + \cdots \right)\left( \delta_\sigma{}^\nu + \omega_\sigma{}^\nu + \cdots \right) \\& = g^{\mu \nu} + g^{\mu \sigma} \omega_\sigma{}^\nu + g^{\rho \nu} \omega_\rho{}^\mu + O(\omega^2) \\& = g^{\mu \nu} + \omega^{\mu\nu} + \omega^{\nu \mu} + O(\omega^2) \end{aligned} \end{equation} and so: \begin{equation} \omega^{\mu\nu} = - \omega^{\nu \mu} \end{equation} Thus, the matrix $\omega$ is a $4 \times 4$ antisymmetric matrix, which corresponds to $6$ independent parameters (i.e. the $3$ parameters corresponding to boosts and the $3$ parameters corresponding to rotations).