From a path integral point of view, the Ward-Takahashi identity is a straightforward consequence of the fundamental theorem of calculus.  

Let $\delta$ be a vector field on the space $\mathcal{F}$ of fields which expresses the action of the group of gauge symmetries.  Suppose that $d\phi$ is a measure which is invariant under these gauge symmetries.  The fundamental theorem of calculus tells us that the integral of the total derivative

$\int_{\mathcal{F}} \delta(g) d\phi = 0,$

for any $g$ such that the contribution from the boundary of $\mathcal{F}$ is $0$.

In the special case where $g$ is the product $\mathcal{O}e^{-S}$ of a gauge-invariant observable and the exponential of a gauge-invariant action, we get a constraint on the path integral (expectation with respect to $e^{-S(\phi)}d\phi$):

$0 = \int_{\mathcal{F}} \delta(\mathcal{O}(\phi) e^{-S(\phi)})d\phi = \int_{\mathcal{F}}\bigl[\delta(\mathcal{O}) - \mathcal{O} \delta(S) \bigr]e^{-S(\phi)}d\phi = \langle \delta \mathcal{O}\rangle - \langle \mathcal{O} \delta(S) \rangle.$

If we use the fact that the variation of the action is the spacetime integral of the divergence of the current $\delta(S) = \int_\Sigma \nabla \dot{} J$, then we get the usual Ward-Takahashi identity:

$\langle \delta \mathcal{O}\rangle = \int_\Sigma \langle \mathcal{O}  \nabla \dot{} J\rangle.$

Everything I've written here is only obviously true for finite-dimensional integrals.  However, any path integral should be arbitrarily well approximated by such finite-dimensional integrals, so one can at least hope to transport the final result, even if the steps themselves can not be carried over.