Note: In the case of gauge symmetry, the degeneracy of the vacuum under gauge transformations leads to topologically inequivalent vacua characterized by the winding number of the gauge fields, in which case the lagrangian in the path integral has a term which indeed depends on which (theta) vacuum you choose. However here we will consider a vacuum degeneracy due to a global symmetry.
Proof that defining equation for path integral is independent upon choice of vacuum (it is assumed that you agree that the in and out vacuum are the same, because otherwise the matrix element vanishes), for simplicity let's consider a one parameter continuous symmetry:
First remember that for a general S-matrix element
\begin{align*}
\left\langle \beta^+ \right| T\{ \ldots \} \left| \alpha^- \right\rangle &= \int \prod_{\tau,\vec{x},m}d \phi_m (\vec{x},\tau)\{ \ldots\} \text{exp }\left(i\int_{-\infty}^{+\infty}d\tau L[\phi(\vec{x},\tau),\dot{\phi}(\vec{x},\tau)]\right) \\
&\qquad\qquad \times\left\langle \beta^+ \right| \left.\phi(+\infty);+\infty\right\rangle \left\langle \phi(-\infty);-\infty\right|\left. \alpha^- \right\rangle
\end{align*}
The $\pm$ superscript is for out and in state respectively. We parametrize the vacuum by $\theta$
$$
e^{Q\theta}\left| \Omega,0\right\rangle = \left| \Omega,\theta\right\rangle
$$
Where $Q(\pi,\phi)$ is the generator of this continuous global symmetry. Fist, chose the in and out states to be the $\theta=0$ vacuum.
Then you can easily prove that
$$
\left\langle \Omega^{\pm},0\right| \left.\phi(\pm\infty);\pm\infty\right\rangle \propto \text{exp} \left( -\frac{1}{2}\int d^3 xd ^3 y\mathcal{E}(x,y)\phi(x)\phi(y) \right)
$$
Where $\mathcal{E}(x,y) = \mathcal{E}(x-y)$ is the Fourier transform of the free energy as a function of the momentum.
Then if we chose the in and out states to be the vacuum with small $\theta>0$
\begin{align*}
\left\langle \Omega^{\pm},\theta\right| \left.\phi(\pm\infty);\pm\infty\right\rangle &\propto \left(1 + \theta Q\left[\phi,\frac{\delta}{\delta \phi}\right] \right)\text{exp} \left( -\frac{1}{2}\int d^3 xd ^3 y\mathcal{E}(x,y)\phi(x)\phi(y) \right) \\
&\approx \text{exp} \left( -\frac{1 + \theta}{2}\int d^3 xd ^3 y\mathcal{E}(x,y)\phi(x)\phi(y) \right)
\end{align*}
And we deduce that in both cases $\theta = 0$ and $\theta $ slightly larger, the term
$$
\left\langle \Omega^+, \theta \right| \left.\phi(+\infty);+\infty\right\rangle \left\langle \phi(-\infty);-\infty\right|\left. \Omega^- ,\theta \right\rangle
$$
Will have the only purpose of providing the $i\epsilon$ term, and there for $\theta$ is irrelevant.