I would like an example of the following situation, or a proof that no such example exists.
$\textbf{Situation}$: Two connective (EDIT: I'm fine with dropping this condition) spectra $X$ and $Y$ such that the spaces $\Omega^\infty\Sigma^nX$ and $\Omega^\infty\Sigma^nY$ are equivalent (as spaces) for all $n$, but as spectra, $X$ and $Y$ are $\textit{not}$ equivalent. It was pointed out to me by Maxime that another way to phrase things is that $X$ and $Y$ are two (grouplike) $E_\infty$ spaces that are equivalent as $E_n$-spaces for all finite $n$ but not as $E_\infty$-spaces.
I think (as Charles pointed out in the comments, this is wrong) it would be sufficient to find an example of the following: An infinite loop space $A$ (with terms $A_0=\Omega A_1,A_1,…$ and structure maps $a_i:A_i\rightarrow \Omega A_{i+1}$ such that $a_0$ is the identity) and an equivalence $f:A_0\rightarrow A_0$ (of spaces) such that for any (infinite loop) equivalence $\phi:A\rightarrow A$ with 0th component $\phi_0:A_0\rightarrow A_0$, the composite $\phi_0\circ f$ is not homotopic to the component of any infinite loop map. Because then we can consider the infinite loop space $B$ with the same terms $B_i=A_i$ and the same structure maps $b_i=a_i$ for $i>0$ but with the $b_0$ changed to $f$. The condition on $f$ implies that there is no way to fill in the 0th component of any attempted equivalence between $A$ and $B$.
The problem is that due to my insanely low supply of brain cells the only infinite loop spaces I can really come to grips with are Eilenberg-Maclane, and among Eilenberg-Maclane spectra such an example can't exist since those determined by their homotopy groups…
Best Answer
For this we can use a swindle-type technique.
Let $B = \bigoplus_{n=2}^\infty H\Bbb Z/2$, and $A = \bigoplus_{n=2}^\infty \Sigma^{-n} H\Bbb Z/2$. We can construct maps $A \to B$ by specifying their effect on each summand of $A$.
Define $X = cofib(f)$ and $Y = cofib(g)$. Since these sums start at $n=2$, $B$ is the connective cover of both $X$ and $Y$ by the long exact sequence in homotopy. I claim that $X$ and $Y$ have equivalent connective covers $\tau_{\geq m} X$ and $\tau_{\geq m} Y$ for all $m < 0$, and that they are inequivalent spectra. This suffices to give an example for your question, since taking the associated infinite loop spaces factors through some connective cover.
First let's show that they have equivalent covers. Note that $\tau_{\geq m+1} X$ is the cofiber of a map $$ \bigoplus_{2 \leq n \leq -m} \Sigma^{-n} H\Bbb Z/2 \xrightarrow{\tau_{\geq m} f} \bigoplus_{n=2}^\infty H\Bbb Z/2 $$ and similarly for $\tau_{\geq m+1} Y$. However, the maps $\tau_{\geq m} f$ and $\tau_{\geq m} g$ are equivalent maps in the homotopy category: both are the the same finite list of Steenrod squares summed with countably many extra summands of $H\Bbb Z/2$ on the target. Thus, these connective covers are equivalent.
Next, let's show that $X$ and $Y$ are not equivalent. To see this, first note that $Y$ has $H\Bbb Z/2$ as a summand: the map $g: A \to B$ completely misses the summand $H\Bbb Z/2$ and allows us to split it off. However, the cofiber sequence defining $X$ gives an exact sequence $$ [X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \to [A, H\Bbb Z/2] $$ which is more explicitly $$ [X,H\Bbb Z/2] \to \prod_{n=2}^\infty H^0(H\Bbb Z/2) \xrightarrow{\prod Sq^n} \prod_{n=2}^\infty H^n(H\Bbb Z/2). $$ This second map is injective, and so by exactness the restriction $$ [X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \cong Hom(\pi_0 X, \Bbb Z/2) $$ must be trivial. Therefore, $X$ can't have $H\Bbb Z/2$ as a summand and thus can't be equivalent to $Y$.