When calculating the tree propagator it's not matter whether You field is abelian or not.
You can obtain the expression of the propagator in the following equivalent way:
Operator approach. You start from the definition of a propagator:
$$
\tag 1 D_{\mu\nu}(x - y) = \langle 0| T(W_{\mu}^{\dagger}(x)W_{\nu}(y))|0\rangle
$$
After straightforward calculation (see, for example, Weinberg's QFT Vol. 1) it can be shown that this expression is nothing but
$$
D_{\mu\nu}(x - y) \simeq \int \frac{d^{4}p}{(2\pi)^{4}} e^{ip(x-y)}D_{\mu\nu}(p), \quad D_{\mu\nu}(p) = \frac{\sum_{\sigma}\epsilon^{\dagger}_{\sigma,\mu}(p)\epsilon_{\sigma, \nu}(p)}{p^{2} - m^{2}}
$$
The expression $\sum_{\sigma}\epsilon^{\dagger}_{\sigma,\mu}(p)\epsilon_{\sigma, \nu}(p)$ is the polarization sum rule, which can be derived from the lagrangian.
Path integral approach. In this approach
$$
(1) \equiv \frac{\delta^{2}Z[J] }{\delta W^{\mu,\dagger}(x)\delta W^{\nu}(y)}\bigg|_{J = 0},
$$
where
$$
Z[J] = \frac{1}{\int DW_{\mu}^{\dagger}DW_{\mu}e^{iS[W_{\mu},W_{\mu}^{\dagger}]}}\int DW^{\dagger}_{\mu}DW_{\mu} e^{iS[W_{\mu},W^{\dagger}_{\mu}] + i\int d^{4}xW_{\mu}J^{\mu,\dagger} - i\int d^{4}xW_{\mu}^{\dagger}J^{\mu}}
$$
is generating functional. In our case it is Gaussian integral, and it can be shown that for bose variables $W_{\mu}, W_{\mu}^{\dagger}$ it is equal to
$$
Z[J] = e^{i\int d^{4}x J_{\mu}^{\dagger}(x)\big(K^{\mu\nu}(x-y)\big)^{-1}J_{\nu}(y)},
$$
where $K^{\mu\nu}(x)$ is the action quadratic form operator
$$
S = \int d^{4}x W_{\mu}^{\dagger}(x)\big(K^{\mu\nu}(x)\big)^{-1}W_{\nu}(x)
$$
So, $D_{\mu\nu}(p) = K_{\mu\nu}^{-1}(x)$.
It is not matter in which approach - path integral or operator formalism - you work. They give the same answer, by definition
