Quantum Anomalies and Quantum Symmetries
Comments to the question (v2):
Traditionally, the classical action $S$ sits in the Boltzmann factor $\exp\left[\frac{i}{\hbar} S\right]$ behind an inverse power of $\hbar$ in the path integral, while the path integral measure is independent of $\hbar$. In the conventional way of counting, we say that the Jacobian $J$ from the path integral measure is a one-loop effect proportional to $\hbar$, while the variation of the classical action $S$ is tree-level, i.e. independent of $\hbar$. Anyway, the upshot is, that in a usual setting, the two variations carry different $\hbar$-orders, and cannot cancel.
However, in principle one may introduce a quantum action $$\tag{A} W(\hbar)~=~S+ \sum_{n=1}^{\infty}\hbar^n M_n~=~ S+\hbar M_1+{\cal O}(\hbar^2)$$ with quantum terms. Then the Boltzmann factor becomes $$\tag{B} \exp\left[\frac{i}{\hbar} W\right]~=~\exp\left[\frac{i}{\hbar} S\right]e^{iM_1}(1+{\cal O}(\hbar)),$$ so that a cancellation may formally take place between the $M_1$-action factor and the path integral measure.
It seems appropriate to mention that such cancellation is the main idea behind the quantum master equation (QME) $$ \frac{1}{2}(W,W)~=~i\hbar\Delta W\tag{QME}$$ in the Batalin-Vilkovisky formalism. The lhs. and rhs. of the above QME are associated with the action and Jacobian, respectively, leading to a (generalized) BRST symmetry of the path integral.
Nevertheless, in practice in a local QFT, the BV operator (aka. the odd Laplacian) $\Delta$ is singular object. The QME is typically only satisfied if both sides of the QME are zero separately, i.e. the action and the measure parts cancel separately in practical applications.
References:
I.A. Batalin & G.A. Vilkovisky, Gauge Algebra and Quantization, Phys. Lett. B 102 (1981) 27–31.
W. Troost, P. van Nieuwenhuizen & A. Van Proeyen, Anomalies and the Batalin-Vilkovisky lagrangian formalism, Nucl. Phys. B333 (1990) 727.
nLab.