"Universal" differential identities

Let's work in two dimensions for notational simplicity. We claim that there is no non-trivial polynomial identity of the form $$ P( f, f_x, f_y, f_{xx}, f_{xy}, \dots ) = 0$$ relating some finite number of derivatives of $f$, for any smooth $f$. Suppose for contradiction that this is the case. Taking a functional derivative (i.e. replacing $f$ by $f+ \varepsilon g$ for arbitrary smooth $g$, and then differentiating at $\varepsilon=0$, one obtains a linearised (in $g$) identity of the form $$ P_f(f,f_x,\dots) g + P_{f_x}(f,f_x,\dots) g_x + \dots = 0.$$ Now, at any point in space, the values of finitely many of the various derivatives $g, g_x, g_y, \dots$ are completely unconstrained (as can be seen by taking $g$ to be a finite Taylor series around that point). So the only way the above identity can hold for all $f$ and $g$ is if one has $$ P_f(f,f_x,\dots) = 0$$ $$ P_{f_x}(f,f_x,\dots) = 0$$ etc.. By induction on the degree of $P$, this can only happen if $P_f, P_{f_x}, \dots$ vanish identically, so that $P$ is constant, hence zero, so the identity is trivial.

Edit: after the edits of OP this seems to be an incomplete answer.

Are there relations in this ring which are not consequences of the fundamental relation (1)?.

If by this ring you mean the subring of Maps($C^\infty(\mathbb{R}^n),C^\infty(\mathbb{R}^n)$) generated by partial derivatives (where I consider composition as multiplication), then the answer is, no. Indeed, the map from $\mathbb{Z}[v_1,\ldots,v_n]$ (the free commutative ring in $n$ generators, aka polynomials with integer coefficients), to Maps($C^\infty(\mathbb{R}^n),C^\infty(\mathbb{R}^n)$), sending each generator $v_i$ to the partial derivative operator $D_i$ is injective, since from

$\sum_{\alpha\in\mathbb{N}^n} c_\alpha D^\alpha (f)=0$ for all functions $f\in C^\infty(\mathbb{R}^n)$,

we can conclude that all coefficients $c_\alpha\in \mathbb{Z}$ are zero, by applying it to monomial functions $f(x_1,\ldots,x_n)=x^\alpha$. (Here $\alpha =(\alpha_1,\ldots,\alpha_n)$ is a multi-index and $x^\alpha = x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ etc.)

Consider the ring of $\mathcal{D}(n,k)$ of polynomial differential operators of $k$ smooth functions $f_1(x_1,\ldots,x_n), \ldots, f_k(x_1,\ldots, x_n)$ of $n$ variables. That is every element of $P \in \mathcal{D}(n,k)$ is a polynomials $P(x;f,\partial f, \partial^2 f, \ldots)$ in $f$ and finitely many of its partial derivatives, with coefficients being smooth functions of $x$. All of your examples fit into this category. Actually, $\mathcal{D}(n,k)$ is not only a ring, but also a differential ring; it has $n$ independent derivations commuting, $\partial_i \partial_j = \partial_j \partial_i$, and satisfying the Leibniz rule, $\partial_i(PQ) = (\partial_i P) Q + P (\partial_i Q)$. Also, it has the differential subring $\mathcal{D}(n,0) \subset \mathcal{D}(n,k)$ consisting of smooth functions of $n$ variables, with the usual action of partial derivatives on them.

Now, one can make the following statement:

The differential ring $\mathcal{D}(n,k) \cong \mathcal{D}(n,0)[f_1,\ldots, f_k]$ is a differential ring extension of $\mathcal{D}(n,0)$ freely generated by adjoining the generators $f_1, \ldots, f_k$.

Does this statement answer your question? The only identities used in the construction of the extension to $\mathcal{D}(n,k)$ from $\mathcal{D}(n,0)$ are those satisfied by all differential rings, namely the commutativity of the derivations (which just partial derivatives, in this case) and the Leibniz rule.