= Solution
For a finite-dimensional integral, $u_i e^{b\cdot u}=\partial_{b_i}e^{b\cdot u}$. Expanding a polynomial interaction $V(u)$ formally and interchanging each finite-order derivative with the regulated integral gives
$$
\int d^nu\,e^{-u^TAu/2+b\cdot u+V(u)}
=e^{V(\partial_b)}I_0(b),
$$
where $I_0$ is the free <Gaussian integral>. This is a perturbation-series identity; it does not assert convergence of the full expansion for an arbitrary potential.
For the Minkowski <functional integral>, each field insertion is generated by
$$
\phi(x)e^{i\int J\phi}
=\frac1i\frac{\delta}{\delta J(x)}e^{i\int J\phi}.
$$
Expand $e^{iS_I[\phi]}$, replace its field factors by these <functional derivatives>, and take the derivatives outside the free integral. The result is <polynomial interactions as source differential operators>:
$$
\boxed{Z[J]=\exp\left[i\int d^dx\,
\mathcal L_I\left(\frac1i\frac{\delta}{\delta J(x)}\right)\right]Z_0[J].}
$$
If the <field interaction term> contains derivatives of the field, the corresponding spacetime derivatives act on the source-insertion operators as well. For the scalar polynomial interactions the operator in the exponent is
$$
-i\int d^dx\left[
\frac{\lambda_3}{3!}\left(\frac1i\frac{\delta}{\delta J(x)}\right)^3
+\frac{\lambda_4}{4!}\left(\frac1i\frac{\delta}{\delta J(x)}\right)^4
+\frac{\lambda_6}{6!}\left(\frac1i\frac{\delta}{\delta J(x)}\right)^6\right].
$$
The free quadratic source exponential then supplies all <Wick contractions>, reproducing the <Feynman rules>. For normalized <correlation functions>, divide the resulting interacting functional by $Z[0]$; the interaction operator changes this denominator too, so it cannot simply be ignored.
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