EDIT: I'll leave the old answer up, see below. In the meantime, Maxime Ramzi and I have thought this through and came up with a fun general argument.
Claim. Let $L: \mathrm{Sp}\to \mathrm{Sp}$ be a Bousfield localisation (not necessarily smashing!) which takes bounded below spectra to bounded below spectra. Then it is given (for bounded below $X$) by one of the following:
- $LX = X[P^{-1}]$ for $P$ a set of primes.
- $LX = \prod_{p\notin P} X^\wedge_p$ for $P$ a set of primes.
To see this, first observe that $L$ takes $\mathbb{Z}$-modules to $\mathbb{Z}$-modules. This is because $\mathrm{Mod}_{\mathrm{Sp}}(\mathbb{Z})$ admits a description as Lawvere theory, namely as finite-product preserving functors $\mathrm{Fun}^{\Pi}(\mathrm{Latt}^{\mathrm{op}},\mathrm{Sp})$, which is functorial in exact functors.
Let $C$ denote the functor given by the cofiber $X\to LX \to CX$. Then the above shows that for an abelian group $A$, $LA$ and $CA$ are both generalized Eilenberg-MacLane. So they split as $\bigoplus \pi_nLA[n]$ and $\bigoplus \pi_nCA[n]$. So $\pi_n CA$ is $L$-acyclic and $\pi_nLA$ $L$-local. By the long exact sequence, for $n\notin\{0,1\}$, we have $\pi_nCA = \pi_nLA$, so they must be $0$. It follows directly that $LA$ and $CA$ are $0$-connective and $1$-truncated.
By Postnikov induction, this implies that for any $a$-connective and $b$-truncated $X$, $LX$ is $a$-connective and $b+1$-truncated.
We also have that there exists $n$ such that $LX$ is $-n$-connective for all $X$, where $n$ does not depend on $X$. Otherwise we would find a sequence of connective $X_n$ such that $LX_n$ is not $-n$-connective, and then $L(\bigoplus X_n)$ could not be bounded below at all, since it admits all $LX_n$ as retracts.
The sequence $\tau_{\geq n} X \to X \to \tau_{\leq n} X$ now combines with the two statements above to prove that $LX$ is connective for connective $X$. That is: A bounded below localisation automatically
preserves connectivity.
Next, call an abelian group $L$-acyclic if $LA=0$, and $L$-local if $LA=A$. If all homotopy groups of $X$ are $L$-acyclic, $X$ is $L$-acyclic, by Postnikov induction, and analogously for $L$-locality. We prove the converse. Assume $X$ is $L$-acyclic, then in the sequence
$$
L\tau_{\geq n} X \to 0 \to L\tau_{\leq n-1}X
$$
the left term is $n$-connective and the right term is $n$-truncated, so both have to be $0$. This shows that all truncations of bounded below $L$-acyclic spectra are $L$-acyclic. The same argument using the functor $C$ proves the locality statement.
A localisation $L$ preserving bounded-below spectra is thus classified by the $L$-acyclic abelian groups. These form a localising subcategory of $\mathcal{D}(\mathbb{Z})$, which are classified by their support and generated by $\mathbb{F}_p$ for a set of primes $p\in P$, and possibly $\mathbb{Q}$. All possible combinations are given by the two alternatives in the claim.
Old answer:
As discussed in the comments, the opposite of your premise seems to be true: Basically all interesting localisations seem to take non-bounded below values.
To substantiate this, let's classify the smashing localisations which take the sphere to a bounded-below object and see that they are exactly given by the ordinary localisations and $0$.
We have $L\mathbb{S} \otimes L\mathbb{S} \simeq L\mathbb{S}$. So the base-change $LK = L\mathbb{S}\otimes K$ for $K$ any ordinary field (we'll use $\mathbb{Q}$ and $\mathbb{F}_p$) satisfies $LK \otimes_K LK \simeq LK$. As derived modules over a field always split as a sum of shifts of $K$, we either have that $LK$ is zero, or that base-change along $K\to LK$ is conservative. Combining with the fact $LK\otimes_K LK$, we learn that for any field, the unit map $K\to LK$ is an equivalence, or $LK\simeq 0$.
Now consider $\mathbb{Z}\to L\mathbb{Z}$. As we saw above, for any $p$ we either have $L\mathbb{F}_p=0$ or that $\mathbb{F}_p\simeq L\mathbb{F}_p$ is an equivalence. In the latter case, we learn that $L\mathbb{Z}\to L\mathbb{F}_p$ is surjective on $\pi_0$, so multiplication by $p$ on $\pi_{-1} L\mathbb{Z}$ is injective. If $L\mathbb{F}_p=0$, it is even bijective. Since $L\mathbb{Q}$ is either $0$ or $\mathbb{Q}$, and we may write it as colimit of $L\mathbb{Z}$ along multiplication with all integers, we learn that $\pi_{-1}L\mathbb{Z}=0$. So $L\mathbb{Z}$ is connective. The fact that $L\mathbb{F}_p=0$ or $\mathbb{F}_p$ also tells us that $\pi_0L\mathbb{Z}$ is $p$-torsion free. So if $L\mathbb{Z}\neq 0$, $L\mathbb{Q}$ has to be $\mathbb{Q}$ and cannot be $0$.
Let $P$ be the set of primes for which $L\mathbb{F}_p=0$, the map $\mathbb{Z}\to L\mathbb{Z}$ factors through a map $\mathbb{Z}[P^{-1}]\to L\mathbb{Z}$. Unless $L\mathbb{Z}=0$, this map is now an equivalence after base-change to any $\mathbb{F}_p$ or $\mathbb{Q}$, so by Hurewicz, it is an equivalence.
So far we have not used the assumption that $L\mathbb{S}$ is bounded below at all, we have just proved that all smashing localisations over $\mathbb{Z}$ are either $0$ or ordinary localisations. This comes in now, when relating $\mathbb{Z}$ and $\mathbb{S}$:
Recall that for bounded-below spectra, being $\mathbb{F}_p$-acyclic is the same as being trivial mod $p$ (by Hurewicz), so for any prime $p$ with $L\mathbb{F}_p=0$, $L\mathbb{S}/p=0$ as well. The map $\mathbb{S}\to L\mathbb{S}$ thus factors through $\mathbb{S}[P^{-1}]\to L\mathbb{S}$. This is now a map of bounded below spectra which is an equivalence after tensoring with $\mathbb{Z}$, if $L\mathbb{Z}\neq 0$. So $L\mathbb{S}\simeq \mathbb{S}[P^{-1}]$. If $L\mathbb{Z}=0$, we instead directly observe that $L\mathbb{S}\otimes \mathbb{Z} \simeq 0$, so $L\mathbb{S}=0$ by Hurewicz.
To summarize: Smashing localisations with $L\mathbb{S}$ bounded below are either ordinary localisations or trivial.
