No, it is not cyclic.

For each of the rings, the semigroup of endomorphisms consists of one element of degree $n$ for each positive integer $n$ plus one or two elements of degree $0$. For a positive integer, this is given by the unique product of Frobenius lifts of that degree, i.e. in $R(\mathbb C)$ by $x \mapsto x^n$ and in $R(SL_2)$ by the $n$'th (normalized) Chebyshev polynomial of the first kind. That these are endomorphisms follows from the characterization of torsion-free lambda-rings in terms of commuting lifts of Frobenius.

The degree $0$ elements are $x\to 0, x\to 1$ for $R(\mathbb C)$ and $x\to 2$ for $R(SL_2)$.

For maps from one ring to the other, the only choices are degree $0$, i.e. $x\to 2$ for $R(SL_2) \to R(\mathbb C)$ and $x\to 0$ or $x\to 1$ for $R(\mathbb C) \to R(SL_2)$.

To calculate these, we can use the following trick: Both rings embed into $\mathbb Z[t,t^{-1}]$, with the lambda-ring structure arising from viewing it as the representation ring of $\mathbb G_m$, by the embeddings $x\to t$ and $x\to t+t^{-1}$. So it suffices to find the lambda-ring homomorphisms from each ring to $\mathbb Z[t,t^{-1}]$ and then throw away the ones whose image does not lie inside the desired target rings.

Since $\lambda^2(x) = 0 $ for $x$ the standard generator of $R(\mathbb C)$, any homomorphism $R(\mathbb C) \to \mathbb Z[t,t^{-1}]$ must send $x$ to an element $y$ with $\lambda^2(y)=0$. It's easy to see that such an element must have leading coefficient $1$ and no coefficients below the leading coefficient, i.e. must be a momomial $t^n$ for $n\in \mathbb Z, or have no leading coefficients at all, since otherwise the leading term of $\lambda^2(y)$ would be nonvanishing. Then one can check that $x\to 0$ and $x \to t^n$ are in fact homomorphisms. The image of these homomorphisms is contained in $R(\mathbb C)$ if and only if $n\geq 0$, and contained in $R(SL_2)$ only if $n=0$ or $x$ is sent to $0$.

Similarly, $R(SL_2(\mathbb C))$ is generated by an element $x$ with $\lambda^2(x)=1$, so $x$ must be sent to an element $y$ with $\lambda^2(y)=1$. Such an element must have leading coefficient $1$ or $2$, with the second case only in degree $0$, or otherwise the leading term of $\lambda^2(y)$ is something other than $1$. In the first case, the leading term of $\lambda^2(y)$ comes from the next term beyond the leading term of $y$, so we must have $y = t^n + t^{-n} + $ lower order terms. and in the second case we have $y = 2+$ lower order terms. Symmetrically, we must have $y= t^n + t^{-n} + $ higher order terms or $y = 2+$ higher order terms, and this is only possible if $y= t^n+ t^{-n}$ for $n\geq 0$ (with $n=0$ encapsulating the $y=2$ case). One checks that $x \mapsto t^n + t^{-n}$ indeed gives a homomorphism for all $n$, and then observes that its image lies in $R(SL_2)$ for all $n$ and lies in $R(\mathbb C)$ only if $n=0$.

No, it is not cyclic.

For each of the rings, the semigroup of endomorphisms consists of one element of degree $n$ for each positive integer $n$ plus one or two elements of degree $0$. For a positive integer, this is given by the unique product of Frobenius lifts of that degree, i.e. in $R(\mathbb C)$ by $x \mapsto x^n$ and in $R(SL_2)$ by the $n$'th (normalized) Chebyshev polynomial of the first kind. That these are endomorphisms follows from the characterization of torsion-free lambda-rings in terms of commuting lifts of Frobenius.

The degree $0$ elements are $x\to 0, x\to 1$ for $R(\mathbb C)$ and $x\to 2$ for $R(SL_2)$.

For maps from one ring to the other, the only choices are degree $0$, i.e. $x\to 2$ for $R(SL_2) \to R(\mathbb C)$ and $x\to 0$ or $x\to 1$ for $R(\mathbb C) \to R(SL_2)$.

To calculate these, we can use the following trick: Both rings embed into $\mathbb Z[t,t^{-1}]$, with the lambda-ring structure arising from viewing it as the representation ring of $\mathbb G_m$, by the embeddings $x\to t$ and $x\to t+t^{-1}$. So it suffices to find the lambda-ring homomorphisms from each ring to $\mathbb Z[t,t^{-1}]$ and then throw away the ones whose image does not lie inside the desired target rings.

Since $\lambda^2(x) = 0 $ for $x$ the standard generator of $R(\mathbb C)$, any homomorphism $R(\mathbb C) \to \mathbb Z[t,t^{-1}]$ must send $x$ to an element $y$ with $\lambda^2(y)=0$. It's easy to see that such an element must have leading coefficient $1$ and no coefficients below the leading coefficient, i.e. must be a momomial $t^n$ for $n\in \mathbb Z$, or have no leading coefficients at all, since otherwise the leading term of $\lambda^2(y)$ would be nonvanishing. Then one can check that $x\to 0$ and $x \to t^n$ are in fact homomorphisms. The image of these homomorphisms is contained in $R(\mathbb C)$ if and only if $n\geq 0$, and contained in $R(SL_2)$ only if $n=0$ or $x$ is sent to $0$.

Similarly, $R(SL_2(\mathbb C))$ is generated by an element $x$ with $\lambda^2(x)=1$, so $x$ must be sent to an element $y$ with $\lambda^2(y)=1$. Such an element must have leading coefficient $1$ or $2$, with the second case only in degree $0$, or otherwise the leading term of $\lambda^2(y)$ is something other than $1$. In the first case, the leading term of $\lambda^2(y)$ comes from the next term beyond the leading term of $y$, so we must have $y = t^n + t^{-n} + $ lower order terms. and in the second case we have $y = 2+$ lower order terms. Symmetrically, we must have $y= t^n + t^{-n} + $ higher order terms or $y = 2+$ higher order terms, and this is only possible if $y= t^n+ t^{-n}$ for $n\geq 0$ (with $n=0$ encapsulating the $y=2$ case). One checks that $x \mapsto t^n + t^{-n}$ indeed gives a homomorphism for all $n$, and then observes that its image lies in $R(SL_2)$ for all $n$ and lies in $R(\mathbb C)$ only if $n=0$.