Linearly distributive categories are the models of the fragment of linear logic consisting of the two connectors $\otimes$, $⅋$ and the two multiplicative units $\top$ for $\otimes$ and $\bot$ for $⅋$. This is classical multiplicative linear logic minus the negation.
Models of classical muliplicative linear logic are $*$-autonomous categories. If you have a $*$-autonomous category with tensor $\otimes$, monoidal unit $\top$ and duals noted $A^{\perp}$, you can define $A ⅋ B := (A^{\perp} \otimes B^{\perp})^{\perp}$ and $\bot = \top^{\perp}$ and it gives you a linearly distributive category.
$*$-autonomous categories axiomatize multiplicative linear logic in terms of tensor and negation. Linearly distributive categories doesn't completely axiomatize classical multiplicative linear logic in terms of tensor and par because it doesn't give you the negation. If you add objects $A^{\perp}$ and maps $$\top \rightarrow A ⅋ A^{\perp}$$ $$A \otimes A^{\perp} \rightarrow \bot$$ satisfying appropriate coherence conditions, you recover a $*$-autonomous category.
I think it explains the second and third sentences. Now, the first.
From one proof of each of this sequents in linear logic: $$A,B\vdash C,D$$ $$D,E \vdash F$$ you obtain a proof of this sequent $$A,B,E \vdash C,F$$ by using the Gentzen cut rule. In a linearly distributive category we can perform this step in terms of (ordinary) categorical composition like this:
The two proofs correspond to two morphisms:
$$
f:A \otimes B \rightarrow C ⅋ D
$$
$$
g:D \otimes E \rightarrow F
$$
We can form
$$
f \otimes E: A \otimes B \otimes E \rightarrow (C⅋D) \otimes E
$$
$$
C ⅋ G: C ⅋ (D \otimes E) \rightarrow C ⅋ F
$$
To compose them you need an intermediate morphism
$$
(C ⅋ D)\otimes E \rightarrow C ⅋ (D \otimes E)
$$
between the two and it's exactly one of the four distributive natural transformations defining (non symmetric in the paper) linearly distributive categories, page 52 in the paper of Cockett and Seely.
If you compose the three morphisms, you obtain one of the appropriate type
$$
A \otimes B \otimes E \rightarrow C ⅋ F
$$
corresponding to the proof obtained by using the cut rule in linear logic.
You can surely find informations in Categorical Semantics of Linear Logic by Paul-André Melliès in addition to the nLab (for example, you have the page star-autonomous category).
I don't really know about
the adjunction $Theories \rightleftarrows Categories$ described in the nLab-article syntactic category.
I don't think this is really any more satisfying than the example of a one-object category, but you can start with your favorite additive category and then delete objects so that biproducts don't exist. For example, you can consider the category of vector spaces of dimension $\le d$ for some fixed positive integer $d$.
The distinction between additive and pre-additive categories is really not that important, for the following reason. Every pre-additive category $C$ admits an additive completion $\widetilde{C}$ in which one formally adjoins the zero object and finite biproducts, and this additive completion is unique in a very strong sense: morphisms between any two objects in $\widetilde{C}$ are completely determined as matrices of morphisms between their direct summands in $C$, so there's no choice whatsoever about how to adjoin the zero object and finite biproducts. Moreover, the inclusion $C \to \widetilde{C}$ preserves any biproducts which already exist in $C$ (note that the analogous statement for, say, the Yoneda embedding is quite false). $\widetilde{C}$ also has the left adjoint property that the category of functors from $C$ to any additive category $D$ is equivalent to the category of functors from $\widetilde{C}$ to $D$, so e.g. their $\text{Ab}$-valued presheaf categories are equivalent.
Applying this construction to the one-object pre-additive category with endomorphisms a ring $R$ produces the category of finite free $R$-modules. So the difference between these two categories is very mild. Similarly, applying it to the category of finite free $R$-modules of rank $\le d$ also produces the category of finite free $R$-modules.
Among other things, this shows that every example of a pre-additive category is obtained from an additive category by deleting some objects.
Best Answer
Instead of comparing groups and rings, one needs to compare monoids and rings:
We see from this that the multiplication in a monoid needs to be compared to the multiplication in a ring. The addition in rings has no corresponding structure in monoids.
Every ring has an underlying additive group, and some monoids happen to be groups. But that doesn’t matter for the above observation: the multiplication in a group is a special case of the multiplication in a monoid, and therefore needs to be compared to the multiplication in a ring (and not the addition). The addition in rings has no corresponding structure in groups.