The algebraic de Rham complex

I have found a more simple construction. In fact, there is a universal property of $\Omega^1_{A/R}$ as an $R$-module: It is the cokernel of the $A$-linear (and therefore $R$-linear) map $A \otimes_R A \otimes_R A \to A \otimes_R A$ defined by $a \otimes b \otimes c \mapsto ab \otimes c - b \otimes ac - a \otimes bc$. This alternative construction seems to be quite unknown(?), see also here. The differential is given by $d(a)=[a \otimes 1]$. We exactly mod out the Leibniz rule $d(a \cdot b)=a \cdot d(b)+b \cdot d(a)$, and the $c$ gives us the (right) $A$-module structure.

Now consider the $R$-linear map $A \otimes_R A \to \Omega^2_{A/R},~ a \otimes b \mapsto d(b) \wedge d(a)$. It extends to $\Omega^1_{A/R}$ since $${\small d(c) \wedge d(ab) - d(ac) \wedge d(b) - d(bc) \wedge d(a)}$$ $${\small =a \cdot d(c) \wedge d(b) +b \cdot d(c) \wedge d(a)-a \cdot d(c) \wedge d(b) - c \cdot d(a) \wedge d(b)-b \cdot d(c) \wedge d(a)-c \cdot d(b) \wedge d(a)=0}.$$

More generally, $\Omega^p_{A/R}$ is an explicit quotient of $A^{\otimes 2p}$, and $A^{\otimes 2p} \to \Omega^{p+1}_{A/R}$ defined by $a_1 \otimes b_1 \otimes \dotsc \otimes a_p \otimes b_p \mapsto d(b_1 \cdot \dotsc \cdot b_p) \wedge d(a_1) \wedge \dotsc \wedge d(a_p)$ preserves the relations (easy calculation as above), so that it extends to $d^p : \Omega^p_{A/R} \to \Omega^{p+1}_{A/R}$.

This construction even generalizes to arbitrary commutative algebra objects in cocomplete symmetric monoidal categories. In the abelian case we can define de Rham cohomology. Details will appear elsewhere ;) (namely my thesis, the section about derivations).

I think there is an easier proof.

Write $A$ as a quotient $B/J$ where $B$ is a polynomial ring over $R$ (not necessarily finite type). Then $\Omega^1_{B/R}$ is free over $B$, and you pointed out, the construction is easy in this case. We have the exact sequence $$ J\stackrel{\delta}{\to} \Omega_{B/R}^1\otimes A \to \Omega_{A/R}^1\to 0$$ which implies that $\Omega_{A/R}^i$ is canonically the quotient of $\Omega_{B/R}^i$ by $J\Omega_{B/R}^i$ and the submodule generated by $d(J)\wedge \Omega_{B/R}^{i-1}$. To define $d^i : \Omega_{A/R}^i \to \Omega_{A/R}^{i+1}$, it is enough to check that $d^i : \Omega_{B/A}\to \Omega_{B/R}^{i+1}$ maps $J\Omega_{B/R}^{i}$ and $d(J)\wedge \Omega_{B/R}^{i-1}$ to the submodule generated by $d(J)\wedge \Omega_{B/R}^{i}$ and $J\Omega_{B/R}^{i+1}$, but this is almost obvious.

It remains to check the condition on $d^{p+q}(\eta\wedge \omega)$. This trivially follows from the same condition on the $\Omega_{B/R}^i$'s.