Exact solution of a class of interacting QFTs Consider a Euclidean quantum field theory, whose action only has derivative terms. For example, for a simple scalar field, consider
$$S[\phi]=\int\mathrm{d}^d{x}\ \mathrm{P}_{2n}[\partial\phi], \tag{1}$$
where $\mathrm{P}_{2n}[\partial\phi]$ is an arbitrary polynomial of degree $2n$. The $(\partial\phi)^2$ term is what we'd call the usual kinetic terms, and the rest are increasingly more irrelevant terms. For well-definedness of some of the calculations below I will assume that the coefficient of the highest order term is positive. If necessary I'm also willing to drop all odd terms, though I don't think it's a huge issue.
Let's now consider the path integral for this theory
$$Z = \int \mathcal{D}\phi\ \exp(-S[\phi]),$$
and do the change of variables $A=\partial\phi$$(*)$. Then we have
$$Z = \int \mathcal{D}A\ \frac{\mathcal{D}\phi}{\mathcal{D}A}\ \exp(-S[A]).$$
But it holds that
$$\frac{\mathcal{D}\phi}{\mathcal{D}A} = \frac{1}{\mathrm{det}\,\partial} = \frac{1}{\sqrt{\mathrm{det}\,\partial^2}},$$
up to, zero-mode issues and possibly, regularization issues, which are not hugely important at this point (I could be more careful if necessary to deal with those, but I don't think it's important for now). The rest of the path integral is now a path integral of a theory with no derivatives, so it reduces to an infinite product of regular convergent integrals,
$$\int \mathcal{D}A \exp\!\left(-\int\mathrm{d}^d x\; \mathrm{P}_{2n}[A]\right) = \prod_x \left(\int \mathrm{d}A(x) \exp\!\left(-\mathrm{P}_{2n}[A(x)]\right)\right)=:Z_0,$$
where I possibly need to regularize the infinite product in the end.
All in all we end up with, what we'd call a one-loop determinant $Z_1:=\left(\det\partial^2\right)^{-1/2}$ and $Z_0$, having, thus, fully solved the theory:
$$Z=Z_0\; Z_1\tag{2}$$
getting a manifestly finite answer.
Generalizing, one could imagine replacing $\mathrm{P}_{2n}[\partial \phi]$ by an arbitrary positive definite function of $\partial \phi$. One cute choice would be $\sqrt{\mathrm{det}(\partial\phi\otimes\partial\phi)}$ giving the Nambu-Goto action. All of the above arguments also work in this case.
I strongly suspect there is something wrong with my argumentation! For one, this gives - albeit just conceptually - an explicit prescription to quantize the Nambu-Goto action, which, is famously difficult (and hence we resort to quantizing the Polyakov action in string theory). But also, just looking at the action (1) (with $n\to\infty$), it includes an infinite tower of ever-more irrelevant terms, hence one would assume it is UV-divergent, but the result (2) is finite, hence the theory is UV-finite.
Where is the catch?

$(*)$ I've suppressed indices to not clutter the notation, but everything should work when restoring indices. Alternatively one should look at it through a differential form lense
 A: I guess I'll write this as an answer instead of a series of comments, though it may not fully answer the question. My original comment considered a special case of the following model.
Consider the following (non-quantum) lattice model of $N$ sites and lattice spacing $a$ for a set of variables $\phi_i \in \mathbb{R}$, $i = 1, \dots, N$:
$$p(\phi) \propto \exp\left( - \sum_{i=1}^N\left\{ \frac{\rho a}{2} \left(\frac{\phi_{i} - \phi_{i-1}}{a} \right)^2 + \frac{k a}{2} \phi_i^2\right\}\right).$$
For now, let me consider a system with periodic boundary conditions, $\phi_{i+N} = \phi_i$.
In the continuum limit $N \rightarrow \infty$, $a \rightarrow 0$, $Na = L$, we take $i/N \rightarrow x \in (0, L]$, $\phi_i \rightarrow \phi(x)$, and $(\phi_{i+1} - \phi_i)/a \rightarrow \phi'(x)$. i.e., this is a 1-dimensional model. The resulting action would be
$$S[\phi] = \int_0^L dx~\left\{ \frac{\rho}{2} \left(\phi'(x) \right)^2 + \frac{k}{2} \phi(x)^2\right\},$$
with $\phi(x + L) = \phi(x)$. If $k = 0$, this would correspond to $P_{2}(\partial \phi)$ in OP's notation. I'll just focus on this quadratic model, since it should be sufficient to illustrate my point.
Let me return to the lattice model for now. In my original comment I considered $k = 0$, $\rho = 1$, $a = 1$, and $N = 3$, and pointed out that if one integrates $\phi = (\phi_1,\phi_2,\phi_3)$ over $\mathbb{R}^3$ the probability density is not normalizable when $k = 0$. You could try to save this by restricting the values the $\phi$'s take to some finite range, but placing bounds on the $\phi$'s is like adding some very nonlinear terms to the action in order to drive the probably to zero when $\phi$ exceeds the desired bounds, and the resulting action cannot really be thought of as a function of just $\partial \phi$ in this case.
We can see this in the general case by attempting to change variables to the lattice derivative. Let $A_i = (\phi_i - \phi_{i-1})/a$ for $i = 2, \dots, N$ and $A_1 = (\phi_1 - \phi_N)/a$ (due to the periodic boundary conditions). Then, $\partial A_i /\partial \phi_j = (\delta_{ij} - \delta_{i-1,j})/a$, $i = 2, \dots, N$ and $\partial A_1/\partial \phi_j = (\delta_{1,j} - \delta_{N,j})/a$. In matrix form this Jacobian looks like
$$\frac{\partial A}{\partial \phi} = \frac{1}{a} \begin{bmatrix} 1 & 0 & 0 &\dots & -1 \\ -1 & 1 & 0 & \dots & 0 \\ 0 & -1 & 1 & \dots & 0 \\ \vdots & 
 & & & \vdots \\ 0 & 0 & 0 & \dots & 1\end{bmatrix};$$
i.e., ones on the diagonal, $-1$ on the lower diagonal, except in the last column in which there is a $-1$ in the top row. This matrix has determinant $0$, as can be seen, e.g., by sequentially adding each row to the one above it--- eventually you will find that the top two rows differ by a factor of $-1$. Thus, this transformation is singular for the $1d$ lattice ring, and the model is not well-defined when $k = 0$.
It's also instructive to try to calculate $\partial \phi/\partial A$ instead of $\partial A/\partial \phi$ (either one works if the jacobian determinant is non-zero). To do so we need to solve for $\phi_i$ in terms of the $A_i$. We can rewrite our definition of $A_i$ as $\phi_{i+1} = \phi_1 + a A_i$, which is a recursion relation with solution $\phi_{i} = \phi_1 + a \sum_{j=2}^i A_j$. If we set $i = N$ we get $0 = \phi_1 - \phi_N + a\sum_{j=2}^N A_j = a \sum_{j=1}^N A_j$, which shows that the $A_i$'s are not independent --- which manifests as the singularity in $\partial A/\partial \phi$.
This singularity results from the periodic boundary condition. We might try to save the model, at least in 1 spatial dimension, by considering a chain instead.
But, to do that we need to decide what happens to the $((\phi_i - \phi_{i-1})/a)^2$ term for $i = 1$. One option is to just leave it out. We could then define $A_i = (\phi_i - \phi_{i-1})/a$ as before, but only for $i = 2, \dots, N$, and $\phi_1$ would have to be left on its own and not part of the transformation. The matrix $\partial A/\partial \phi$ would then have $1$'s along the diagonal and $-1$ on the lower diagonal, but no pesky $-1$ in the top right corner. The matrix has determinant $1/a^{N-1}$, so the change of variables is ok. But then the normalization integral becomes
$$\int_{-\infty}^\infty d\phi_1 d\phi_2 \dots d\phi_N ~\exp\left( - \sum_{i=2}^N\left\{ \frac{\rho a}{2} \left(\frac{\phi_{i} - \phi_{i-1}}{a} \right)^2\right\}\right) = a^{N-1}\int_{-\infty}^\infty d\phi_1 \int_{-\infty}^\infty dA_2 \dots dA_N ~ \exp\left( - \sum_{i=1}^N\left\{ \frac{\rho a}{2} \left(A_i\right)^2\right\}\right)$$
which is not normalizable because the action does not depend on $\phi_1$, which must still be integrated over. In order to render the integral finite, we must supply a boundary condition to fix $\phi_1$, such as by multiplying the exponential by a delta function $\delta(\phi_1 - c)$ for the desired boundary value $c$. More generally, one could let the value of $\phi_1$ fluctuate, by demanding it be distributed according to some normalizable distribution. Either way, this explicitly adds $\phi_1$ dependence to the action, so it is not purely a function of derivatives of the fields.
Note that these results do not depend on the fact that the model is only Gaussian in the $A_i$; adding higher order terms will not fix the fact that the transformation is singular or not normalizable unless some explicit dependence on at least one $\phi_i$ is introduced.
tl;dr: a lattice model that would formally limit to a path integral (on a bounded spatial domain) with only derivative terms is not normalizable/singular unless one introduces constraints on the individual fields. According to Qmechanic's comment, this is not improved in higher dimensions.
