Yes, this is correct, except that the plethystic monoidal product refers not to a monoidal product on the category of $R$-rings, but to a monoidal product on the category of co-$R$-ring objects in the category of $R$-rings. (What one might call an $R$-biring.) Or, it could also refer to a right action

$$\odot \colon R\text{-Ring} \times R\text{-Biring} \to R\text{-Ring}$$

of this monoidal category, acting on the category of $R$-rings (we describe this below).

A good way to think about this is that an $R$-biring is an $R$-ring $S$ that carries enough extra structure so as to endow each hom-set $\hom(S, A)$ with an $R$-ring structure, in a way that is natural in $R$-rings $A$. Thus, an $R$-biring is essentially the same thing as an endofunctor

$$R\text{-Ring} \to R\text{-Ring}$$

of the form $\hom(S, -)$. Yes, the hom-functor $\hom(S, -)$ goes from $R\text{-Ring}$ to $Set$, but the idea is that we also have natural maps representing addition

$$a: \hom(S, A) \times \hom(S, A) \to \hom(S, A),$$

multiplication, etc. on each hom-set, to make each such hom-set an $R$-ring. Using the fact that we have a natural isomorphism

$$\hom(S, -) \times \hom(S, -) \cong \hom(S \otimes_R S, -)$$

we see that the natural map $a$ must (by an application of the Yoneda lemma) be represented by an $R$-ring map of the form $\alpha: S \to S \otimes_R S$. Similarly, the natural map $m$ for multiplication is represented by another $R$-ring map $\mu: S \to S \otimes_R S$, and so on for each defining operation in the theory of $R$-rings.

Then, if $S$ and $T$ are two birings, the plethystic product $S \odot T$ is defined to be the biring that represents the composite

$$R\text{-Ring} \stackrel{\hom(T, -)}{\to} R\text{-Ring} \stackrel{\hom(S, -)}{\to} R\text{-Ring}.$$

(Yes, any two such endofunctors represented by birings compose to give another such endofunctor.)

How does this help us? Let's take the polynomial algebra $S = R[x_1, x_2, \ldots, x_n]$., and see it as a biring. (Here $\alpha \colon S \to S \otimes_R S$ is identified with the $R$-ring homomorphism

$$R[x_1, \ldots x_n] \to R[x_1, \ldots, x_n, y_1, \ldots, y_n]$$

taking $x_i$ to $x_i + y_i$. The other co-operations are similar.) We have $\hom(S, A) \cong A^n$, so clearly the hom-set here does carry natural $R$-ring structure, namely the $R$-ring $A^n$. Thus, the biring $S$ represents the functor

$$R\text{-Ring} \stackrel{n^{th}\text{-power}}{\to} R\text{-Ring}$$

and the composite of two such functors is by multiplying powers. Thus we indeed have a natural identification of birings

$$R[X] \odot R[Y] \cong R[X \times Y]$$

where $X$ and $Y$ are finite sets with $m$ and $n$ elements. This is completely general, and $m$ and $n$ do not need to be finite.

How does one construct the plethystic product explicitly from this point of view? The general point is that endofunctors of the form $\hom(S, -)$ are continuous (preserve limits) and have left adjoints. The left adjoint is denoted $- \odot S: R\text{-Ring} \to R\text{-Ring}$ (this gives the right monoidal category action mentioned above). To calculate it, use the fact that left adjoints preserve colimits, and hence any presentation of an $R$-ring $A$ as a coequalizer of the form

$$R[Y] \stackrel{\to}{\to} R[X] \to A.$$

Thus, applying $- \odot S$ to this sequence, we obtain a coequalizer diagram

$$S^{\otimes Y} \stackrel{\to}{\to} S^{\otimes X} \to A \odot S.$$

Here $S^{\otimes X}$ is simply the $X$-fold coproduct of copies of $S$ in the category of $R$-rings. This gives an explicit construction of the ring $A \odot S$. If $A$ is a biring, this construction lifts to a biring structure on $A \odot S$; the details are easy, but I will omit them.

The monoidal unit for $\odot$ is the polynomial algebra in one generator, since this represents the identity endofunctor. Now suppose we have (following your notation) an $R$-ring map $p \colon R[x] \to A$ representing an element of $A$, and an $R$-*biring* map $q: R[x] \to S$. Then we can indeed form

$$R[x] \cong R[x] \odot R[x] \stackrel{p \odot q}{\to} A \odot S$$

and this represents an element of $A \odot S$, which by abuse of notation we denote by $p \odot q$. Taking the case $A = R[x_1, \ldots, x_n]$ and following the definitions above, one easily calculates $p \odot q$ to be the element represented by

$$R[x] \stackrel{p}{\to} R[x_1, \ldots, x_n] \cong R[x]^{\otimes n} \stackrel{q^{\otimes n}}{\to} S^{\otimes n} \cong A \odot S$$

and from here it is not difficult to assure yourself that your formula for $p \odot q$ is indeed correct.