It seems to me that you got derailed by a rather unhelpful discussion in Schwartz. There's a lot going on here which I will briefly outline.
The first thing I want to briefly discus is the relevence of representations of the LORENTZ group vs. representations of the POINCARE group.
Lorentz group: Representations are finite dimensional, given by two half integers $(j_1, j_2)$, and not unitary. Spacetime translation is not included. Representations of this group are representations of local field operators $\hat \phi_i(x)$. Under a Lorentz transformation, the indices of $\hat \phi_i(0$) will get mixed together while the spacetime point $0$ will be fixed. Two simple examples include the spin $0$ representation, where $i$ is just $0$, and the $(1/2, 1/2)$ representation, where we instead use the symbols $\hat A_\mu$ and $\mu = 0... 3$. The fact that these finite dimensional representations (which just juggle around the field indices) are not unitary is totally uninteresting because there isn't a natural inner product of local operators, like there is with states, so who cares that they're not unitary!?
Poincare group: Representations are infinite dimensional, unitary, given (by Wigner's theorem) by either a mass $m^2$ and a half integer spin, or in the $m = 0$ case by a half integer helicity. These correspond to particle states. The fact they they are unitary is important because you want the group action to preserve the natural inner product on states. The fact they are infinite dimensional is not really unexpected, the same thing happens in non relativistic QM! For instance, in the spin $0$ case, single particle states are given by a momentum $| p \rangle$. Under a Poincare transformation,
$$
U(\Lambda, a) | p \rangle = e^{- i p a} | \Lambda p \rangle.
$$
The fact that you need to include a state $| \Lambda p \rangle$ for each Lorentz transformation $\Lambda$, this representation is infinite dimensional because your basis of single particle states is spanned by an infinite number of $p$'s. But this is also the case in non relativistic QM, that you have a basis spanned by an infinite number of momenta, and isn't special to quantum field theory, so it's not really that big a deal. It's true that the Poincare group is non compact because of boosts, but its also non compact because of translations, which is why QM is infinite dimensional (there's an infinite number of positions).
Relationship between the two: You can create single particle states by acting on the vacuum with field operators. In the spin $0$ case a single particle state is given by
$$
| p \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \hat \phi(\vec x) | 0 \rangle
$$
and in the spin $1$ case by
$$
| p, \varepsilon \rangle = \int d^3 x e^{- i \vec p \cdot \vec x} \varepsilon^\mu \hat A_\mu(\vec x) | 0 \rangle
$$
(modulo maybe some integration measure I forgot.) Here $\varepsilon^\mu$ is a polarization vector satisfying $\varepsilon^\mu p_\mu = 0$. Because single particle states are super positions of states like, say, $\hat A_\mu | 0 \rangle$, the spins of the physical particle states can be found by restricting oneself to the rotation part of the Lorentz group, $U(R)$, and just using the standard angular momentum addition formula, where the irreducible representations of the $j_1 \otimes j_2$ representation is $j_1 +j_2 \oplus j_1 + j_2 - 1 \oplus \ldots \oplus |j_1 - j_2|$. So in the $(1/2, 1/2)$ representation of the Lorentz group, we have $1/2 \otimes 1/2 = 1 \oplus 0$. However, the non physical $0$ representation is projected out by the condition $\varepsilon^\mu p_\mu$, getting rid of the negative norm states, yadda yadda yadda. The point is that $j_1 \otimes j_2$ give the possible particle spins, but some of them will be non physical.
The main point, though, is that representations of the Lorentz group are used for talking about field operators, while representations of the Poincare group are used for talking about particle states. Once you realize this, it's completely reasonable that representations of the Lorentz group are not unitary (why should they be?) and that representations of the Poincare group are not finite dimensional (why should they be?).
Best Answer
To every Lie algebra $\mathfrak{g}$ there is an associated simply-connected Lie group $G_s$. By the general correspondence of Lie groups and algebras, every representation of $G_s$ induces one of $\mathfrak{g}$ and vice versa, regardless of surjectivity of the exponential map. Hence a representation of $G_s$ is completely reducible if and only if the corresponding representation of $\mathfrak{g}$ is.
For any connected Lie group $G_c$ with Lie algebra $\mathfrak{g}$, this is still true: If we have a representation $V_r$ of $G_c$ that has a subrepresentation $W_r\subset V_r$ of $G_c$ and $\mathfrak{g}$ is semi-simple, then we get $V_r = W_r\oplus W'_r$ for some other representation $W'_r$ of $\mathfrak{g}$. But $W'_r$ must be a subrepresentation of $G_c$ also, since you can always use the exponential map locally to get the representation of $G_c$ from the representation of $\mathfrak{g}$ - very similar to the general proof of the algebra-group correspondence, surjectivity is not needed here either, only connectedness so that you can "shift" the exponential map along paths in the group.
For a disconnected Lie group $G_d$ with Lie algebra $\mathfrak{g}$, we have that all connected components are diffeomorphic to the identity component $G_0$ and there is an exact sequence $$ 1 \to G_0 \to G \to \pi_0(G)=G/G_0\to 1. \tag{1}$$ If this sequence splits, we have $G = \pi_0(G)\rtimes G_0$ where $\rtimes$ is a semi-direct product. If $\pi_0(G)$ is additionally Abelian, Mackey's theory of induced representations tells us that every representation of $G$ comes from representations of the factors $\pi_0(G)$ and $G_0$ and every pair of irreducible representations of the factors defines an irreducible representation of $G$. Since Maschke's theorem tells us that finite group have only completely reducible finite-dimensional representations, this means that a Lie group for which (1) splits and for which $\pi_0(G)$ is Abelian has completely reducible representations if $\mathfrak{g}$ is semi-simple.
Applying this to the Lorentz group, we have $\pi_0(G) = \mathbb{Z}_2\times\mathbb{Z}_2$ as the group of space parity and time reversal. It is Abelian, and one can show that the Lorentz group is indeed the semi-direct product of this and its connected component by explicitly writing out its matrix structure. Therefore, all finite-dimensional representations of the Lorentz group are completely reducible, and its irreducible representations can be given by an irreducible representation of the proper orthochronous Lorentz group $(V,\rho)$ + an irreducible representation of time reversal (which under the tensor product of the induced representation becomes just some operator on $V$ that squares to the identity) + an irreducible representation of space parity (likewise).