One of the default examples of ordinary graded commutative rings is the polynomial ring $\mathbf Z[t]$. Let us first examine the analogue of that, and then see where else that leads!
1. $S$-grading on $S\{t\}$
For the sake of clarity, allow me to denote the underlying infinite loop space (equivalently: grouplike $\mathbb E_\infty$-space) of the sphere spectrum $S$ by $\Omega^\infty(S)$.
Recall that, by the Barrat-Priddy-Quillen Theorem, the ininite loop space $\Omega^\infty(S)$ can be described as the group completion of the groupoid of finite sets and isomorphisms $\mathcal F\mathrm{in}^\simeq$. Consider the constant functor $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$ with value $S$ - since the latter is the monoidal unit, this is a symmetric monoidal functor. Left Kan extension along the canonical map $\mathcal F\mathrm{in}^\simeq \to (\mathcal F\mathrm{in}^\simeq)^\mathrm{gp}\simeq \Omega^\infty(S)$ produces a symmetric monoidal functor $\Omega^\infty(S)\to \mathrm{Sp}$, and as such exhibits an $S$-graded $\mathbb E_\infty$-ring.
But which one? To figure out which one, note that the passage to the "underlying $\mathbb E_\infty$-ring" of an $S$-graded $\mathbb E_\infty$-ring is given by passage to the colimit. Thus the underlying $\mathbb E_\infty$-ring, which we have just adorned with an $S$-grading, is
$$
\varinjlim_{\Omega^\infty(S)}\mathrm{LKan}^{\Omega^\infty(S)}_{\mathcal{F}\mathrm{in}^\simeq}(S)\simeq \varinjlim_{\mathcal F\mathrm{in}^\simeq}S,
$$
where we used that the left Kan extension of a functor does not change its colimit. Now we can use the explicit description of the groupoid of finite sets $\mathcal F\mathrm{in}^\simeq \simeq \coprod_{n\ge 0}\mathrm B\Sigma_n$ to obtain
$$
\varinjlim_{\mathcal F\mathrm{in}^\simeq} S\simeq \bigoplus_{n\ge 0}S_{h\Sigma_n}.
$$
We may recognize this as the free $\mathbb E_\infty$-ring on a single generator $S\{t\}$, corresponding in terms of spectral algebraic geometry to the smooth affine line $\mathbf A^1$ (as opposed to the flat affine line $\mathbf A^1_\flat = \mathrm{Spec}(S[t])$ for the polynomial $\mathbb E_\infty$-ring $S[t]\simeq \bigoplus_{n\ge 0}S$).
2. $S$-grading on symmetric algebras
The previous example can be easily generalized by observing that $\mathcal F\mathrm{in}^\simeq$ is the free $\mathbb E_\infty$-space ( = symmetric monoidal $\infty$-groupoid) on a single generator. That means that a symmetric monoidal functor $\mathcal F\mathrm{in}^\simeq \to \mathrm{Sp}$ (which always factors through the maximal subgroupoid $\mathrm{Sp}^\simeq\subseteq\mathrm{Sp}$) is equivalent to the data of a spectrum $M\in \mathrm{Sp}$ (the image of the singleton set). The functor is given by sending a finite set $I$ to the smash product $M^{\otimes I}$, and is evidently symmetric monoidal. The same Kan extension game as before now gives rise to an $S$-graded $\mathbb E_\infty$-ring spectrum, this time with underlying $\mathbb E_\infty$-ring
$$
\mathrm{Sym}^*(M)\simeq \bigoplus_{n\ge 0}M^{\otimes n}_{h\Sigma_n},
$$
the free $\mathbb E_\infty$-ring generated by $M$. We recover the prior situation by setting $M=S$. For $M = S^{\oplus n}$, we obtain an $S$-grading on $S\{t_1, \ldots, t_n\}$, corresponding to the spectral-algebro-geometric smooth affine $n$-space $\mathbf A^n$.
3. $S$-grading on $S\{t^{\pm 1}\}$
Another way to generalize the example of the $S$-grading on $S\{t\}$ is to take directly the constant functor $\Omega^\infty(S)\to \mathrm{Sp}$ with value $S$, instead of starting with a constant functor on $\mathcal F\mathrm{in}^\simeq$ and Kan-extending it. This produces a perfectly good $S$-graded $\mathbb E_\infty$-ring, which let us denote $S\{t^{\pm 1}\}$.
From the group-completion relationship between $\Omega^\infty(S)$ and $\mathcal F\mathrm{in}^\simeq$, it may be deduced that $S\{t^{\pm 1}\}$ and $S\{t\}$ are related in terms of $\mathbb E_\infty$-ring localization as $S\{t^{\pm 1}\}\simeq S\{t\}[t^{-1}]$, justifying our notation. Here we are localizing $S\{t\}$ at the element $t\in \mathbf Z[t] = \pi_0(S\{t\})$. In terms of spectral algebraic geometry, this is encoding the spectral scheme $\mathrm{GL}_1$, the smooth punctured line.
4. Remark on non-negative grading
In algebraic geometry, we often prefer to think about non-negatively graded commutative rings than graded commutative rings. Just as the latter are equivalent to lax symmetric monoidal functor $\mathbf Z\to\mathrm{Ab}$, so are the former equivalent to lax symmetric monoidal functors $\mathbf Z_{\ge 0}\to \mathrm{Ab}$.
By analogy, the "non-negatively $S$-graded $\mathbb E_\infty$-rings" are lax symmetric monoidal functors $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$. Indeed, just as $\mathbf Z_{\ge 0}$ is the free commutative monoid on one generator, so is $\mathcal F\mathrm{in}^\simeq$ the free $\mathbb E_\infty$-space on one generator. That is the reason why we were encountering such functors above (and Kan extending them along group completion, as we would to view as $\mathbf Z_{\ge 0}$-grading as a special case of a $\mathbf Z$-grading).
5. Some actual "non-tautological" examples though
So far, a not-completely-unreasonable complaint might be that all the examples of $S$-graded $\mathbb E_\infty$-rings were sort of tautological.
For a very non-tautological example, see the main result of this paper of Hadrian Heine. It shows that there exists an $S$-graded $\mathbb E_\infty$-ring spectrum such that its $S$-graded modules are equivalent to the $\infty$-category of cellular motivic spectra. In fact, much more is proved: this situation is very common, and under some not-too-harsh compact-dualizable-generation assumptions, a symmetric monoidal stable $\infty$-category will be equivalent to $S$-graded modules over some $S$-graded $\mathbb E_\infty$-ring. So you may just as well view this result as a wellspring of potentially interesting examples of $S$-gradings "occurring in nature"! :)
No, this does not work in the non-commutative case. In general we have $HH(A)=A\otimes_{A\otimes A^{\mathrm{op}}} A$, and this is only a $k$-module, not an algebra. If $A$ is commutative, the tensor product happens to compute coproducts/pushouts of commutative $k$-algebras, and we have $HH(A)=\operatorname{colim}_{S^1}A=x_!(A)$ since $S^1=*\coprod_{*\sqcup *}*$, but in the non-commutative case the tensor product does not have such an interpretation.
To see that the group $S^1$ acts on $HH(A)$ in general, one can use the formalism of factorization homology, $HH(A)=\int_{S^1} A$, which makes the functoriality on $BS^1$ apparent. A more classical approach is to use the fact that the usual "Hochschild complex" extends to a cyclic $k$-module (a functor on Connes' cyclic category $\Lambda$), whose geometric realization acquires an action of $S^1$ due to the $\infty$-groupoid completion of $\Lambda$ being $BS^1$. A reference for the latter approach is Appendix B of the article by Nikolaus and Scholze: https://arxiv.org/pdf/1707.01799.pdf
Best Answer
The topological Hochschild cohomology (that I'll denote now THC) makes sense whenever $A$ is at least an $E_1$-algebra. In particular, you can construct THC of an $E_\infty$-algebra. There is a result called Deligne's conjecture but which is now a theorem stating that THC of an $E_1$-algebra is an $E_2$-algebra. In particular, if you take the homology of THC of something, the resulting graded abelian group has a Gerstenhaber algebra structure. If you take homotopy groups, you get a commutative algebra with a degree 1 bracket but I don't think it's going to satisfy the axioms of a Gerstenhaber algebra in general.
Taking the endomorphisms over $A\otimes S^{n-1}$ is a perfectly fine construction called higher THC. It can be defined as soon as $A$ is an $E_{n}$-algebra although the definition is slightly more involved (a good reference is http://www.math.northwestern.edu/~jnkf/writ/cotangentcomplex.pdf). Higher Deligne's conjecture tells you that this higer Hoschild cohomology is an $E_{n+1}$-algebra. In particular taking homology, you get a Gerstenhaber algebra with a bracket of degree $n$.
Note that in the case where $A$ is $E_\infty$, there is a nice construction of higher THC in the following paper of Ginot Tradler and Zeinalian (they restrict to $E_\infty$-algebras in chain complexes but the case of spectra is similar) http://arxiv.org/abs/1205.7056
Edit: I just noticed that you were asking more specifically what THC of $KU$ is. It turns out that the unit map $KU\to F_{KU\wedge KU}(KU,KU)$ is an equivalence. The same is true if you replace $KU$ by $E_n$ (the height $n$ Lubin-Tate spectrum). This remains true for the higher dimensional versions of THC. The unit map $E_n\to F_{S^d\otimes E_n}(E_n,E_n)$ is an equivalence. The reason for this is essentially the fact that $E_n$ is étale aver the $K(n)$-local sphere. You can look at http://geoffroy.horel.org/HHC%20of%20the%20LT%20ring%20spectrum.pdf for more details.