Even for spatial locales this is not true, since there are (sober) topological spaces whose maps to their Stone-Čech compactifications are not injective. An example is the line with two origins.
I don't know a snappy characterization of those locales for which the map $r$ is a monomorphism, and I'm not sure you'll be able to find one, because monomorphisms are not viewed as the "right" notion in the category of locales. Much more attention gets paid to regular monomorphisms (e.g. a sublocale of $L$ is defined to be an isomorphism class of regular monomorphisms into $L$).
And the theorem is that $r$ is a a regular monomorphism if and only if $L$ is a completely regular locale. Similarly, the map from a topological space to its Stone-Čech compactification is a topological embedding if and only if the space is completely regular.
The best reference for this (and all related topics) is probably Johnstone's book Stone Spaces.
Yai0Phah's comment (and your intuition) is right -- $\{\operatorname{Spec}(S_i^{-1}A)\to\operatorname{Spec}A\}$ is a faithfully flat cover of $\operatorname{Spec}A.$ Since this collection is finite, this is the same as saying that $A\to\prod_{i = 1}^n S_i^{-1}A$ is faithfully flat. Let's first prove the algebraic result, and then talk about the geometric implications.
Claim: If $S_1,\dots, S_n$ is a collection of comaximal monoids in $A,$ then the natural map
$$
A\to\prod_{i = 1}^n S_i^{-1}A
$$
is faithfully flat.
Proof: First, notice that map $A\to S_i^{-1}A$ is flat, since any localization is flat (and that each $S_i\subseteq A$ is a multiplicative subset, so we can localize). Then, the direct product is also flat, so all we need to prove is that for any maximal ideal $\mathfrak{m}\subseteq A,$ the tensor product $\left(\prod_{i = 1}^n S_i^{-1}A\right)\otimes_A\kappa(\mathfrak{m})\neq 0$ (see here). Since this is a finite product, it agrees with the direct sum, and so it commutes with the tensor product. Thus, we need to check that for any maximal ideal $\mathfrak{m},$ there exists some $i$ such that $S_i^{-1}A\otimes_A\kappa(\mathfrak{m})$ is nonzero.
Suppose for the sake of contradiction that there is some maximal ideal $\mathfrak{m}\subseteq A$ such that $S_i^{-1}A\otimes_A\kappa(\mathfrak{m}) = 0$ for each $i.$ This means that $S_i^{-1}\kappa(\mathfrak{m}) = 0$ for any $i,$ and the localization of a ring is zero if and only if the multiplicative subset includes $0.$ In this case, this means that for each $i,$ there exists some $s_i\in S_i$ such that the image $s_i + \mathfrak{m}\in\kappa(\mathfrak{m})$ is zero. However, this implies that the ideal generated by these $s_i$ is not the unit ideal, which contradicts comaximality of the $S_i.$ $\square$
In case you're not already aware, to any commutative, unital ring $A$ we may associate a topological space $\operatorname{Spec}A$ whose points are the prime ideals of $A.$ This space naturally comes equipped with a sheaf of rings whose global sections are precisely $A.$ So, the elements of the ring $A$ can be thought of as functions on the space $\operatorname{Spec}A.$
We give $\operatorname{Spec}A$ a topology called the Zariski topology, but the cover above is not a cover in this topology. Instead, this collection is a cover in what's called the fpqc topology, which is a generalization of the classical notion of a topology. We don't talk about open sets in the fpqc topology, but we can still make sense of good covers, and we can talk about sheaves with respect to this topology. The point is, $\{\operatorname{Spec}B\to \operatorname{Spec}A\}$ is an fpqc cover precisely if the ring map $A\to B$ is faithfully flat.
So, the above result is saying that the collection $\{\operatorname{Spec}S_i^{-1}A\to\operatorname{Spec}A\}$ form a nice cover of the space $\operatorname{Spec}A$, although the images of these maps need not be open sets in the Zariski topology on $\operatorname{Spec}A$.
In other words, your theorem is indeed saying that if you can find a solution on each portion of the cover, then those solutions can be glued to give a solution on the entire space. (You might not literally be gluing the local solutions to get a global solution in the proof, but I suspect that is what happens.)
Thinking of the $S_i$ (or any selection of one element from each) as being a partition of unity is good geometric intuition. In fact, I believe this terminology is used in Vakil's notes, among other places (perhaps also Eisenbud's "Commutative Algebra"?).
As for the geometric interpretation of modules, you can indeed think of them as generalizations of vector bundles: see here for a discussion.
Then here is one way to interpret your local-global principle. You are given a certain good (fpqc) cover $\{U_i = \operatorname{Spec}S_i^{-1}A\mid i = 1,\dots, n\}$ of the space $\operatorname{Spec}A,$ and a fixed "vector bundle" $\mathcal{E}$ on $\operatorname{Spec}A.$ Given a system of equations
\begin{align*}
x_1 \sigma_{1,1} + x_2 \sigma_{1,2} + \dots + x_r \sigma_{1,r} &= \tau_1\\
x_1 \sigma_{2,1} + x_2 \sigma_{2,2} + \dots + x_r \sigma_{2,r} &= \tau_2\\
\qquad\vdots\qquad\qquad&\\
x_1 \sigma_{k,1} + x_2 \sigma_{k,2} + \dots + x_r \sigma_{k,r} &= \tau_k\\
\end{align*}
where each of the $\sigma_{i,j},\tau_i$ are global sections of $\mathcal{E},$ can we find functions $f_1,\dots, f_r$ on $\operatorname{Spec}A$ which solve these equations simultaneously? Well, the answer is that if you can find collections of functions locally on each piece of the cover which work, you can glue those functions into functions on the entire space which solve your system.
Best Answer
This is false. For instance, let $A=\mathbb{Z}[x]/(2x)$, let $S_n$ be generated by $x^n$ and let $T$ be generated by $2$. Then it is easy to see that $\bigwedge S_n=\{1\}$ so $0\in S_n\vee T$ for all $n$ but $0\not\in (\bigwedge S_n)\vee T$.
Note that you should not expect this to be the correct construction: localizations of $A$ are not the same thing as submonoids, since different submonoids can give the same localization,. Moreover, open subsets of $\operatorname{Spec} A$ are not the same as localizations of $A$: only localizations with respect to finitely generated (or equivalently, singly generated) submonoids give open sets, and a general open set is a union of open sets that come from localizations. Another reason to be suspicious of your construction is that it uses only the multiplicative structure of $A$, not the additive structure! (And $\operatorname{Spec} A$ as a locale absolutely does depend on the additive structure; for instance if $k$ is a field then $k[x]$ and $k[x,y]$ are multiplicatively isomorphic but their Specs are different.)
From the perspective of algebraic geometry, there is a very simple and natural way to describe $\operatorname{Spec} A$ as a locale: just take the poset of radical ideals in $A$ (by the usual correspondence, radical ideals correspond to closed sets of $\operatorname{Spec} A$ by an inclusion-reversing bijection, and thus to open sets by an inclusion-perserving bijection).
Here's one way to see radical ideals arising naturally from the perspective of localizations. It is enough to consider only localizations with respect to a single element, so instead of arbitrary submonoids of $A$ we will just consider elements of $A$, representing the distinguished open subsets of $\operatorname{Spec}A$. We will represent general open subsets as the collection of all the distinguished open subsets they contain, so our locale $L$ will consist of certain subsets of $A$. That is, we think of an element $S\in L$ as representing the open set $U=\bigcup_{a\in S} D(a)\subseteq\operatorname{Spec} A$, and conversely $S=\{a\in A:D(a)\subseteq U\}$). What properties do these subsets need to satisfy? Well, if $D(a)$ is covered by the sets $D(b)$ for $b\in S$, then $a$ should be in $S$. In other words, if the elements of $S$ generate the unit ideal in the localization $A[a^{-1}]$, then $a$ should be in $S$. Or equivalently, if some power of $a$ is in the ideal of $A$ generated by $S$, then $a\in S$. Now this condition looks familiar: it just says that $S$ must be a radical ideal.
To be fair, this may feel like cheating in that it takes for granted that "$D(a)$ is covered by elements $D(b)$" should mean "the elements $b$ generate the unit ideal in $A[a^{-1}]$". Here is another construction that is very slick and may feel more natural. Take the free frame on $A$ modulo relations that say $1_A$ is $1$, $0_A$ is $0$, $a\wedge b=ab$ for $a,b\in A$, and $a\vee b\geq a+b$. Thinking of $a\in A$ as representing the distinguished open set $D(a)$, this means we are just formally making a frame out of these distinguished open sets, imposing the relation that $D(1)$ is the whole space, $D(0)$ is empty, $D(ab)=D(a)\wedge D(b)$, and $D(a)\vee D(b)$ contains $D(a+b)$. Remarkably, it turns out that this frame is naturally isomorphic to the frame of radical ideals in $A$, by mapping $a\in A$ to the radical ideal generated by $a$.
For more discussion of this construction and other properties of the spectrum from a locale-theoretic perspective, you may be interested in section V.3 of Peter Johnstone's book Stone spaces.