This is a Lie group, and its connected component around the identity results by exponentiating the Lie algebra, whose most general element you already wrote down.
Expanding the generic group element around the identity, you find, as is standard in Lie theory,
$$G(\epsilon, \omega)=e^{i\epsilon_{\mu}P^{\mu}-i\omega_{\mu \nu}M^{\mu\nu}/2}= 1\!\!1 + i\epsilon_{\mu}P^{\mu}-i\omega_{\mu \nu}M^{\mu\nu}/2 + O(\epsilon^2,\omega^2,\epsilon \omega).
$$
Normally, however, one writes the full Poincare transformation, always possible, (by theorem, as your text might review), as a product of a translation T and a proper Lorentz transformation Λ,
$$
U(a,\Lambda)=T(a) \Lambda =e^{ia_{\mu}P^{\mu}} e^{-i {\tilde \omega}_{\mu \nu}M^{\mu\nu}/2},
$$
so that
$$
U( a',\Lambda')U(a,\Lambda)= U(\Lambda' a+ a',\Lambda ' \Lambda).
$$
This is your expression U (left-hand side). Notationally, $e^{ia P} f(x)=f(x+a)$ and $\tilde \omega$ is the "angle" of the "generalized noncompact rotation" matrix $\Lambda= 1\!\!1 -i\tilde\omega\cdot M/2+ O(\tilde\omega^2)$.
You may, always, as per Lie's theorem, compose the two exponential factors into a single exponential that looks like the above G, but with its parameters ε,ω as messy functions of those of each factor exponential in your U, as you saw from the composition law. (You may compute the first few orders for fun. Do you see that $\tilde\omega$ will never be modified, so $\omega=\tilde \omega$? Do you see the first subleading term in the translation parameter, $\tilde \omega \cdot a$?)
