Morphisms of $\mathbb E_l$-rings between $\mathbb E_k$-rings for $l<k$

Let's consider these in characteristic zero, so that we can use differential graded algebras and so that we can model $E_\infty$ things by strictly commutative things and $E_1$ things by associative things.

Let $A$ be the commutative DGA $\Bbb Q[x,y]$ where $x$ and $y$ are in degrees $2n$ and $2m$ respectively, with $dx = dy = 0$. Then the set of maps $A \to B$ in the homotopy category of commutative DGAs can be naturally identified with $H_{2n} B \times H_{2m} B$.

Now let's think of $A$ as an associative DGA; it's no longer semifree, so to compute maps in the homotopy category we first have to find a resolution $\tilde A \to A$. Here's one: the associative $DGA$ given by $$ \tilde A = \Bbb Q\langle x,y,z \rangle $$ where $z$ is in degree $2n+2m=1$ and $dz = xy - yx$. Then $\tilde A$ is semifree, and there is a natural projection map $\tilde A \to A$ that sends $z$ to zero. With some work, one finds that it is a quasi-isomorphism.

As a result, the set of maps $A \to B$ in the homotopy category of $E_1$-algebras are the same as maps $\tilde A \to B$ in the homotopy category of $E_1$-algebras.

Let $B = \Bbb Q[z] / (z^2)$ with $|z| = 2n + 2m + 1$ and $dz = 0$. Then there is only the trivial map $A \to B$ sending $x$ and $y$ to zero, but the projection $\tilde A \to B$ sending $x$, $y$, and $z^2$ to zero is a nontrivial map in the homotopy category. This shows that the forgetful map is not full.

Let $R$ be an $E_\infty$-ring spectrum and write $R\{t\}$ (resp. $R[t]$) for the free $E_\infty$-$R$-algebra (resp. free $E_1$-$R$-algebra) on one generator $t$ (in degree zero). This notation is compatible with your previous question, i.e. $R[t]$ is nothing else than the polynomial $E_\infty$-$R$-algebra on one generator.

The forgetful map $\alpha : Maps_{E_\infty}(R[t], A) \to Maps_{E_1}(R[t], A) \approx \Omega^\infty(A)$ is identified with the canonical map $\varepsilon^* : Maps_{E_\infty}(R[t], A) \to Maps_{E_\infty}(R\{t\}, A) \approx \Omega^\infty(A)$ given by composition with the canonical $E_\infty$-ring homomorphism $\varepsilon : R\{t\} \to R[t]$. The discussion in the previous question implies that whenever $R$ is not a $\mathbf{Q}$-algebra, there exist choices of $A$ such that this map $\varepsilon_R^*$ is not invertible. For those choices, the map $\alpha$ will also fail to be invertible, so in particular there will exist $E_1$-ring homomorphisms $R[t] \to A$ which do not lift to $E_\infty$-ring homomorphisms.