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 thethat 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