The $1$-type of $M\mathbb{Z}/(2^r)$ does have an $E_\infty$-ring structure for $r> 1$. I'm going to show it by using the algebraic models for $1$-truncated connective commutative ring spectra from:

MR2405894 Reviewed Baues, Hans-Joachim; Jibladze, Mamuka; Pirashvili, Teimuraz Third Mac Lane cohomology. Math. Proc. Cambridge Philos. Soc. 144 (2008), no. 2, 337–367.

MR2793446 Reviewed Baues, Hans-Joachim; Muro, Fernando The algebra of secondary homotopy operations in ring spectra. Proc. Lond. Math. Soc. (3) 102 (2011), no. 4, 637–696.

These algebraic models, called $E_\infty$-quadratic pair algebras, give rise to bipermutative categories (see Remarks 5.9, 6.10, and 9.14 in the second paper) which, in turn, can be used to construct 1-truncated connective commutative ring spectra as per:

A. D. Elmendorf and M. A. Mandell, Rings, modules, and algebras in infinite loop space theory, Adv. Math. 205 (2006) 163–228.

I'm sure you can find other proofs, e.g. via an explicit topological construction or lifting the first $k$-invariant to topological André-Quillen cohomology.

Consider the following $E_\infty$-quadratic pair algebra

$$\begin{array}{ccc}
\mathbb{Z}/2\times \mathbb{Z}&\stackrel{\partial}\longrightarrow& \mathbb{Z}\\
\nwarrow^P&&\swarrow_H \\
&\mathbb{Z}&
\end{array}$$

Here,
$$
\begin{array}{rcl}
\partial([a],n)&=&2^rn,\\
P(a)&=&([a],0),\\
H(n)&=&\frac{n(n-1)}{2}.
\end{array}
$$

Moreover, $\mathbb{Z}$ is endowed with the usual product (in a quadratic pair algebra it could have a product which is only right distributive), but it doesn't in this case) and $\mathbb{Z}/2\times \mathbb{Z}$ is endowed with the usual left and right product by elements of $\mathbb{Z}$ (again, the product from the left need not be distributive in general). There could be an additional $\smile_1$ operation that in this case is trivial because all previous products commute.

The $k$-invariant of such a structure is the homomorphism
$$\operatorname{coker}\partial\otimes\mathbb{Z}/2\longrightarrow \ker\partial$$
defined by
$$[a]\otimes[1]\mapsto P(H(2a)-2H(a)).$$

In the previous example, this morphism is $$\mathbb{Z}/2^r\otimes\mathbb{Z}/2\cong \mathbb{Z}/2$$
which coincides with the map
$$\pi_0M\mathbb{Z}/2^r\otimes\pi_1S\longrightarrow \pi_1M\mathbb{Z}/2^r\colon [f]\otimes[g]\mapsto [fg].$$
This is the first $k$-invariant of $M\mathbb{Z}/2^r$.

You may wonder what fails for $r=1$. For the second equation of the second set in Definition 6.1 (in the second of the aforementioned papers) we need $PH\partial=0$. This holds iff $r>1$ since
$$PH\partial([a],n)=\left(\left[\frac{2^rn(2^rn-1)}{2}\right],0\right)=([2^{r-1}n],0).$$

spaces? Not that I have any insight either way, but if you're ultimately interested in something about $\pi$-finite spaces, then maybe truncated Moore spaces are relevant too? $\endgroup$lowerbound for the associativity of $M(p^r)$ when $r$ is large. If I'm reading it right, it says that as $r \to \infty$, the associativity level $n$ of $M(p^r)$ (with $p$ fixed) increases to $\infty$ as well. Soa fortiori, the same is true after truncation. Are you really hoping that $n$ actually reaches $\infty$ at some finite $r$ or something? $\endgroup$