This started as a comment in response to the followup by the OP to Jim's answer, but quickly grew to span several comment blocks, so I'm just expanding it into an answer block.

"I would still be interested in explicit descriptions of the rings $R(E_6)$ and $R(E_7)$. Can they be described as quotients of polynomial rings, like $R(SO(2n))$?"

The short answer is: yes, we can get explicit descriptions, in theory at least.

If $\tilde{G}$ denotes a simply connected group and $G$ its adjoint group, then the irreps of $\tilde{G}$ have a natural grading by the elements of $Z(\tilde{G})$ given by looking at the image of the center in the given irrep. This grading is preserved by taking tensor products, so in particular we can find a set of irreps which span the representation ring of the adjoint group. Irreps of the adjoint group are those one which this grading is trivial; there are then relations among these irreps which can be obtained by looking in the representation ring of $\tilde{G}$.

For example, let $\tilde{G}=SU(3)$ with fundamental weights $w_1$ and $w_2$. In this case an admissible $\mathbb{Z}/3\mathbb{Z}$ grading on an irrep $\pi_{a\cdot w_1+b\cdot w_2}$ (the irrep with highest weight $a\cdot w_1+b\cdot w_2$) of $SU(3)$ is given by $(a−b)\mod 3$. One then sees that all irreps of $G = SU(3)/3$ are generated by $\pi_{3w_1}$, $\pi_{3w_2}$, and $\pi_{w_1+w_2}$; but as $SU(3)/3$ has rank 2, there must be a relation among these three generators to reduce the "dimension" (for whatever the correct notion of dimension of a ring is appropriate here) of its representation ring down by 1; i.e. the representation ring is a quotient of a polynomial ring.

The exact relation can be found by considering tensor products of these three generators and subtracting off appropriate highest weight pieces until one reaches 0. So one starts with $\pi_{w_1+w_2}^{\otimes 3}$ and subtracts off $\pi_{3w_1}\otimes \pi_{3w_2}$ then looks at the highest weights of this representation and subtracts off an appropriate tensor product of the three generator irreps to cancel the highest weights left and repeats until one reaches 0; the resulting polynomial relation has the form (unfortunately I don't recall the exact coefficients $n_i$ and do not currently have access to my Maple worksheets where they are stored):

$\pi_{w_1+w_2}^{\otimes 3} - \pi_{3w_1}\otimes \pi_{3w_2} + n_1\cdot(\pi_{3w_1}\otimes\pi_{w_1+w_2} + \pi_{3w_2}\otimes\pi_{w_1+w_2}) + n_2\cdot \pi_{w_1+w_2}^{\otimes 2} + n_3\cdot(\pi_{3w_1} + \pi_{3w_2}) + n_4\cdot\pi_{w_1+w_2} + n_5\cdot 1 = 0$

Note that this relation is not a problem in $R(SU(3))$ (which is a polynomial ring) since there $\pi_{3w_1}$, $\pi_{3w_2}$, and $\pi_{w_1+w_2}$ are themselves polynomials in the fundamental representations $\pi_{w_1}$ and $\pi_{w_2}$; it is the failure of these two fundamental representations to live in $R(SU(3)/3)$ that forces us to take a 3-element set of generators of $R(SU(3)/3)$ and also forces this 3-element set to have a single relation.

In application to your question, you would need to do a similar construction in $R(E_6)$ and $R(E_7)$: start with the grading function applied to irreps to find a set of irreps of $\tilde{E}_6$ or $\tilde{E}_7$ which generate each subring; then work out relations among these generators by looking at the highest weights and adding/subtracting off appropriate tensor powers of these generators until one arrives at 0. As I don't recall the grading function for these two groups offhand, I couldn't tell you how many generators to expect for each, but once you have all $g_6$ generators for $R(E_6)$, you ought to be able to find exactly $g_6 - 6$ independent relations among them and similarly for the generators of $R(E_7)$.

**Update:** Since it is not difficult to compute the grading function, I went ahead and worked them out so that I could elaborate. The $Z(\tilde{G})$ grade of an irrep $\pi_\omega$ of $\tilde{G}$ is trivial $iff$ $\pi_\omega$ contains the zero weight and the $Z(\tilde{G})$ grades of two irreps are equal iff the highest weight of one is contained in the weights of the other (depending of course on which weight dominates which).

Equivalently if one can reach the zero weight from $\omega$ by subtracting off fundamental roots, or equivalently $\omega$ is a sum of fundamental roots then the grade of $\pi_\omega$ is zero. Another way to think about it is that the grading is really describing the image of the weight in the quotient of the weight lattice modulo
the root lattice.

Since the Cartan matrix expresses weights as sums of fundamental roots, we can use it to solve for this image for an arbitrary weight; simply invert the Cartan matrix and multiply it by the given weight (expressed as a vector over the basis of fundamental weights). If all the entries of the resulting vector are integers, then the weight has trivial grading. If not all entries are integral, then the weight has a nontrivial grade (and there is some choice to this; the choice is restricted by the condition that two weights have the same grade if one dominates the other and the different grading functions can be obtained from one another via a certain action of the outer automorphisms of $\tilde{G}$, so they are all essentially equivalent).

Using the Bourbaki numbering (as found on https://en.wikipedia.org/wiki/Root_system for example), one obtains for $E_6$ that one admissible weight function for $\pi_{\sum_{i=1}^6n_iw_i}$ is given by:

$(n_1+2n_2+n_5+2n_6)\mod 3$

For $E_7$, the weight function for $\pi_{\sum_{i=1}^7n_iw_i}$ is given by:

$(n_1+n_3+n_5)\mod 2$

Now from these we can recover the necessary generators for our adjoint representation rings: we need to find a set of highest weights with trivial grade which integrally span the trivially graded part of the weight lattice (= the root lattice).

For $E_6$ you can check that this set is:

$\{w_3,w_4,3w_1,3w_2,3w_5,3w_6,w_1+w_2,w_1+2w_5,w_1+w_6,w_2+w_5,w_2+2w_6,w_5+w_6\}$

So there are $g_6 = 12$ generators for $R(E_6)$ and so we should be able to find 6 independent polynomial relations among their corresponding irreps which, once known, give the structure of the quotient ring.

For $E_7$ the set of spanning weights is:

$\{w_2,w_4,w_6,w_7,2w_1,2w_3,2w_5,w_1+w_2,w_1+w_3,w_2+w_3\}$

So there are $g_7 = 10$ generators for $R(E_7)$ and so we should be able to find 3 independent polynomial relations among the corresponding irreps.