Great question! This is not a complete answer, but hopefully it gets the ball rolling . . .

**Theorem:** $\mathfrak s_{-}\geq \mathrm{cov}(\mathcal M)$, where $\mathrm{cov}(\mathcal M)$ is the smallest size of a collection of meager sets whose union is $\mathbb R$.

**EDIT:** My first proof of this (which I haven't erased -- you can find it below) uses forcing. I thought it might be nice to include a more direct argument too.

*Second proof (without forcing):* To show that $\mathfrak{s}_{-} \geq \mathrm{cov}(\mathcal M)$, it suffices to show that if $X$ is any set of reals with $|X| < \mathrm{cov}(\mathcal M)$, then $X$ has strong measure zero.

Let $X$ be such a set, and let $\langle \varepsilon_n : n < \omega \rangle$ be a sequence of positive reals. Our goal is to cover $X$ with a sequence $\langle I_n : n < \omega \rangle$ of intervals, where each $I_n$ has length $\varepsilon_n$. Because the $\varepsilon_n$ are arbitrary, this will show that $X$ has strong measure zero.

To achieve our goal, we're going to associate to every point of the Baire space $\omega^\omega$ a sequence of intervals. Given $s \in \omega^\omega$, define $I_n^s$ to be the interval of length $\varepsilon_n$ centered at the $s(n)^{th}$ rational number (where we're imagining that we've fixed some enumeration of $\mathbb Q$ beforehand).

For each $x \in X$, let
$$U_x = \{ s \in \omega^\omega : x \in I^s_n \text{ for some } n \}.$$
It is not too hard to see that $U_x$ is a dense open set for every $x \in X$.

Because $|X| < \mathrm{cov}(\mathcal M)$, the complements of the sets of the form $U_x$ do not cover $\mathbb R$. Thus there is some $s$ that is in $U_x$ for every $x \in X$. This means that $X$ is covered by the sequence of intervals $I^s_1, I^s_2, I^s_3, \dots$, which have the required lengths. **QED**

*First proof (with forcing):* To show that $\mathfrak{s}_{-} \geq \mathrm{cov}(\mathcal M)$, it suffices to show that if $X$ is any set of reals with $|X| < \mathrm{cov}(\mathcal M)$, then $X$ has strong measure zero.

Let $X$ be such a set, and let $\langle \varepsilon_n : n < \omega \rangle$ be a sequence of positive reals. Our goal is to cover $X$ with a sequence $\langle I_n : n < \omega \rangle$ of intervals, where each $I_n$ has length $\varepsilon_n$. Because the $\varepsilon_n$ are arbitrary, this will show that $X$ has strong measure zero.

We will use the following fact about Cohen forcing: *in a Cohen extension, the set of ground model reals has strong measure zero*. This was proved by Martin Goldstern in his first-ever MathOverflow post (although it was known before that). The basic idea of this proof is to show how this fact about Cohen forcing can be converted directly into an inequality between cardinal invariants.

Let $M$ be a transitive model of set theory with $X \in M$, $X \subseteq M$, $\langle \varepsilon_n : n < \omega \rangle \in M$, and with $|M| = |X|$. (Here a "model of set theory" means the usual thing -- only finitely many of the ZFC axioms, but enough to make my argument work!) Such a model $M$ exists by the Löwenheim-Skolem-Tarski Theorem.

Because $|M| < \mathrm{cov}(\mathcal M)$, there is a real $c$ that is Cohen over $M$. This uses the fact that *a real is Cohen over $M$ if and only if it is not in any meager Borel set coded inside of $M$*. Because fewer than $\mathrm{cov}(\mathcal M)$ meager Borel sets are coded in $M$, there is some real missing them all.

By Martin's argument that I linked to earlier, in $M[c]$ we have "the set $X$ is strong measure zero". Now I specifically put the sequence $\langle \varepsilon_n : n < \omega \rangle$ into $M$, which means that this sequence is also in $M[c]$. Thus, because $M[c]$ believes $X$ is strong measure zero and contains this sequence, we have in $M[c]$ that "there is a sequence $I_n$ of intervals of lengths $\varepsilon_n$ covering $X$".

For each $x \in X$, the statement "$x \in I_n$" is absolute. Because $X \subseteq M \subseteq M[c]$, we have, for every $x \in X$, that a statement of this form holds in $M[c]$ (hence in $V$). Thus the sequence of intervals that covers $X$ in $M[c]$ covers $X$. **QED**

I'll add two more observations relating $\mathfrak{s}_{-}$ to cardinals in Cichon's diagram:

$\mathfrak{s}_{-} \leq \mathrm{non}(\mathcal L)$, where $\mathrm{non}(\mathcal L)$ is the smallest size of a non-measurable set.

*Proof:* Every non-measurable set is not strong measure $0$. **QED**

It is consistent that $\mathfrak{s}_{-} < \mathfrak{b}$.

*Proof:* As you mentioned, Laver proved the consistency of $\mathfrak{s}_{-} = \aleph_1$. This holds in what has come to be known as "the Laver model", where it is also known that $\mathfrak{b} = \mathfrak{c} = \aleph_2$. **QED**