Any morphism of $R$-algebras $\varphi : R\{t\} \to R[t]$ is determined up to homotopy by an element of $\pi_0(R[t]) \approx \pi_0(R)[t]$. If $\varphi$ is an equivalence, then this element must be the generator $t$, so we may as well assume $\varphi$ is the canonical map $\varepsilon_R : R\{t\} \to R[t]$.

Claim: Let $R$ be an $E_\infty$-ring spectrum. Then the map $\varepsilon_R$ is invertible if and only if $\pi_0(R)$ is a $\mathbf{Q}$-algebra.

Useful observation: The map $\varepsilon_R$ is compatible with extensions of scalars. Therefore if $\varepsilon_R$ is invertible, then $\varepsilon_{R'}$ is invertible for any $R$-algebra $R'$.

Sufficiency: Suppose that $\pi_0(R)$ is a $\mathbf{Q}$-algebra. The map $\varepsilon_R$ is obtained by extension of scalars along the connective cover $R_{\ge 0} \to R$, so we may assume $R$ is connective.

Since $\pi_0(\varepsilon_R)$ is the identity $\pi_0(R)[t] \to \pi_0(R)[t]$, it suffices to show $\varepsilon_R$ is étale. This is equivalent to $\varepsilon_{\pi_0(R)}$ being étale (cf. Remark 11.2.3.5 in Lurie's "Spectral Algebraic Geometry"). Thus we may assume $R$ is discrete.

If $R$ is a discrete $\mathbf{Q}$-algebra, then the $\infty$-category of connective $E_\infty$-algebras over $R$ is equivalent to that of simplicial commutative $R$-algebras. Under this equivalence, the map $\varepsilon_R$ corresponds to the identity morphism $R[t] \to R[t]$.

Necessity: Suppose $\varepsilon_R$ is an equivalence. Since the map $\varepsilon_R$ is compatible with formation of connective covers, we may replace $R$ by its connective cover. We may also extend scalars along $R \to \pi_0(R)$ to assume $R$ is discrete.

Consider any residue field $R \to k$. If $k$ is of characteristic $p>0$, then $\varepsilon_k$ cannot be invertible. Indeed, it is well-known that $\varepsilon_{\mathbf{F}_p}$ is not invertible, and $\varepsilon_k$ is the extension of scalars along the faithfully flat map $\mathbf{F}_p\{t\} \to k\{t\}$. Thus every residue field $k$ must be of characteristic zero, so $R$ is a $\mathbf{Q}$-algebra.

Any morphism of $R$-algebras $\varphi : R\{t\} \to R[t]$ is determined up to homotopy by an element of $\pi_0(R[t]) \approx \pi_0(R)[t]$. If $\varphi$ is an equivalence, then this element must be the generator $t$, so we may as well assume $\varphi$ is the canonical map $\varepsilon_R : R\{t\} \to R[t]$.

Claim: Let $R$ be an $E_\infty$-ring spectrum. Then the map $\varepsilon_R$ is invertible if and only if $R$ is a $\mathbf{Q}$-algebra.

Useful observations:

1. The map $\varepsilon_R$ is compatible with extensions of scalars. Therefore if $\varepsilon_R$ is invertible, then $\varepsilon_{R'}$ is invertible for any $R$-algebra $R'$.

2. $R$ is a $\mathbf{Q}$-algebra if and only if $\pi_0(R)$ is a $\mathbf{Q}$-algebra (see comments).

Sufficiency: Suppose that $R$ is a $\mathbf{Q}$-algebra. The map $\varepsilon_R$ is obtained by extension of scalars along the connective cover $R_{\ge 0} \to R$, so we may assume $R$ is connective. The $\infty$-category of connective $E_\infty$-algebras over $\mathbf{Q}$ is equivalent to that of simplicial commutative $\mathbf{Q}$-algebras. Under this equivalence, the map $\varepsilon_R$ corresponds to the identity morphism $R[t] \to R[t]$ (where by abuse of notation $R$ also denotes the corresponding simplicial commutative $\mathbf{Q}$-algebra).

Necessity: Suppose $\varepsilon_R$ is an equivalence. Since the map $\varepsilon_R$ is compatible with formation of connective covers, we may replace $R$ by its connective cover. We may also extend scalars along $R \to \pi_0(R)$ to assume $R$ is discrete.

Consider any residue field $R \to k$. If $k$ is of characteristic $p>0$, then $\varepsilon_k$ cannot be invertible. Indeed, it is well-known that $\varepsilon_{\mathbf{F}_p}$ is not invertible, and $\varepsilon_k$ is the extension of scalars along the faithfully flat map $\mathbf{F}_p\{t\} \to k\{t\}$. Thus every residue field $k$ must be of characteristic zero, so $R$ is a $\mathbf{Q}$-algebra.