What is the closure of the Eilenberg-MacLane spectra under limits? under colimits? Every bounded spectrum is in the closure of the Eilenberg MacLane spectra under finite co/limits. Thus every bounded below (resp. above) spectrum is in the closure of the EM spectra under limits (resp. colimits). Let $\mathcal L$ denote the closure of the EM spectra under limits, and let $\mathcal C$ denote the closure of the EM spectra under colimits.
Note that $\mathcal L$ is also closed under (infinite) products, and $\mathcal C$ is also closed under (infinite) coproducts.
Question 1: What is a good description of $\mathcal L$?
Question 2: What is a good description of $\mathcal C$?
Question 3: What is an example of a spectrum in $\mathcal L$ which is not a(n infinite) product of bounded-below spectra?
Question 4: What is an example of a spectrum in $\mathcal C$ which is not a(n infinite) sum of bounded-above spectra?
 A: This is an answer to Questions 3 and 4. Consider the map
$$ H\mathbb Z/2 \xrightarrow{\ldots, Sq^{2^n},\ldots} \prod_{n=1}^\infty \Sigma^{2^n} H\mathbb Z/2.$$
This is a map to a product, where the $n$-th component is the map $Sq^{2^n}\colon H\mathbb Z/2 \to \Sigma^{2^n} H\mathbb Z/2.$ This is a map between products of bounded-above spectra which is not a product of maps between bounded-above spectra. As you suggested in a comment, I believe that the cofiber of this map is an element of $\mathcal C$ that is not a direct sum of bounded above spectra.
It is enough to prove that the fiber of this map is not a direct sum of bounded above spectra. Let $F$ be the fiber. Moreover, for all $m\ge 0$ let $F_m$ be the fiber of the map
$$ H\mathbb Z/2 \xrightarrow{\ldots, Sq^{2^n},\ldots} \prod_{n=1}^m \Sigma^{2^n} H\mathbb Z/2.$$
There is a tower of spectra
$$\cdots F_m \to F_{m-1}\to \cdots $$
and $F$ is the homotopy limit of this tower.
We want to prove that $F$ is not equivalent to a sum of bounded above spectra. I claim that it is enough to prove that none of the maps $F_m\to F_{m-1}$ has a homotopy section. To prove the claim, notice that $F_m$ is the $2^m-1$-th Postnikov section of $F$.
Suppose $F$ splits as a sum (or equivalently a product) of bounded above spectra. Then each Postnikov section of $F$ admits a corresponding splitting of Postnikov sections of factors.
Among the putative factors of $F$ there is one factor, let's call it $X$, that satisfies $\pi_0(X)\cong \mathbb Z/2$. All other factors have $\pi_0=0$. $X$ is bounded above. Let $m$ be the smallest positive integer such that $\pi_{2^m-1}(X)=0$. We have a splitting $F\simeq X\times Y$ for some spectrum $Y$. It induces a splitting of Postnikov sections
$$P_{2^m-1} F\simeq P_{2^m-1}X\times P_{2^m-1}Y.$$
But by our construction, $P_{2^m-1} F=F_m$, $P_{2^m-1}X\simeq F_{m-1}$, and $P_{2^m-1}Y\simeq \Sigma^{2^m-1} H\mathbb Z/2$. It follows that the map $F_m\to F_{m-1}$ has a section.
It remains to prove that none of the maps $F_m\to F_{m-1}$ has a homotopy section. The spectrum $F_m$ is the total fiber of the following square
$$
\begin{array}{ccc}
H\mathbb Z/2 & \xrightarrow{\ldots, Sq^{2^n}, \ldots} & \prod_{n=1}^{m-1} \Sigma^{2^n} H\mathbb Z/2\\
\quad \quad \downarrow Sq^{2^m} & & \downarrow \\
\Sigma^{2^m} H\mathbb Z/2 & \to & *
\end{array}
$$
Note that the fiber of the top map is $F_{m-1}$. There is a fibration sequence
$$F_m \to F_{m-1} \to \Sigma^{2^m} H\mathbb Z/2$$
where the second map is a composition
$$F_{m-1}\to H\mathbb Z/2\xrightarrow{Sq^{2^m}} \Sigma^{2^m}H\mathbb Z/2.$$
If the map $F_m\to F_{m-1}$ has a section, then this composition is null.
But this is not possible, because the kernel of the homomorphism of graded groups
$$\mathcal A\cong [H\mathbb Z/2, H\mathbb Z/2]_*\to [F_{m-1}, H\mathbb Z/2]_*$$
is the left ideal generated by $Sq^2, Sq^4, \ldots, Sq^{2^{m-1}}$, and it does not contain $Sq^{2^m}$.
A context where such a spectrum occurs more or less naturally is the Adams spectal sequence, which in some incarnation probably can be called the Bousfield-Kan spectral sequence. Consider the cobar construction that gives rise to the ASS:
$$H\mathbb Z/2 \Rightarrow H\mathbb Z/2\wedge H\mathbb Z/2 \Rightarrow \cdots H\mathbb Z/2^{\wedge k+1} \cdots$$
(I did not typeset the arrows properly, but it is supposed to be a cosimplicial object in spectra). The second stage of the Tot tower can be identified with the homotopy fiber of a map
$$H\mathbb Z/2 \to H\mathbb Z/2 \wedge \overline{H\mathbb Z/2}$$
where $\overline{H\mathbb Z/2}$ is the cofiber of the map $S\to H\mathbb Z/2$. The spectrum $H\mathbb Z/2 \wedge \overline{H\mathbb Z/2}$ splits as a product of Eilenberg - Mac Lane spectra, and I believe the fiber of the map above is an example of a spectrum in $\mathcal C$ that is not a sum of bounded above spectra. Maybe A.S. had in mind something like this in their comment, but this is just a guess.
Dually, you can construct a spectrum that is in $\mathcal L$ but is not a product of bounded below spectra by taking the fiber of the map
$$\bigoplus_{n=0}^\infty \Sigma^{-2^n} H\mathbb Z/2 \xrightarrow{\ldots, Sq^{2^n}, \ldots} H\mathbb Z/2.$$
