= Solution
Decompose the original <scalar field> as $\Phi=\phi+\widehat\phi$, where $\widetilde\phi(p)$ has support in $|p|\leq\Lambda'$ and $\widetilde{\widehat\phi}(p)$ has support in $\Lambda'<|p|\leq\Lambda$. This is a <momentum-shell decomposition of a scalar field>. The two collections of integration variables are disjoint. Define the lower-scale <Wilsonian effective action> by integrating out the second collection:
$$
\boxed{e^{-S_{\Lambda'}^{\mathrm{eff}}[\phi]}=\int_{\mathrm{shell}}\mathcal D\widehat\phi\,e^{-S_\Lambda^{\mathrm{eff}}[\phi+\widehat\phi]}.}
$$
A field-independent normalization may be retained as a vacuum term or absorbed into the measure. Integrating this identity over the low modes recovers the original <Euclidean path integral>, so it preserves all observables depending only on those modes.
Put $\Delta S=S_\Lambda^{\mathrm{eff}}[\phi+\widehat\phi]-S_\Lambda^{\mathrm{eff}}[\phi]$. The quadratic cross terms integrate to zero: in <Fourier transform> variables each pairs a low momentum with its negative, which cannot be a shell momentum. Expanding the interaction therefore gives
$$
\boxed{\Delta S=\int d^4x\left\{\frac12(\partial\widehat\phi)^2+\frac12m^2\widehat\phi^2+\frac g{24}\left(4\phi^3\widehat\phi+6\phi^2\widehat\phi^2+4\phi\widehat\phi^3+\widehat\phi^4\right)\right\}.}
$$
Factoring $e^{-S_\Lambda^{\mathrm{eff}}[\phi]}$ out of the shell integral and taking minus its logarithm yields
$$
\boxed{S_{\Lambda'}^{\mathrm{eff}}[\phi]=S_\Lambda^{\mathrm{eff}}[\phi]-\log\int_{\mathrm{shell}}\mathcal D\widehat\phi\,e^{-\Delta S[\phi,\widehat\phi]}.}
$$
This definition is a <Wilsonian effective action>, rather than a Legendre transform generating only <one-particle-irreducible Feynman diagrams>.
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