I hope the following can give you some intuition, even though it is not very precise. As a prelude, let us look at the geometric realization of $1$-category $\mathcal{C}$. This is the underlying homotopy type of the nerve $N\mathcal{C}$ of the category. The non-invertible morphisms in $\mathcal{C}$ correspond to the non-invertible (directed) edges of the simplicial set $N\mathcal{C}$, but the underlying homotopy type does not see such direction. Alternatively, the underlying homotopy type is modelled via a Kan fibrant replacement of $N\mathcal{C}$, and Kan complexes have ''inverses'' for all edges. What this means is that the underlying homotopy type of $N\mathcal{C}$ thinks that all morphisms in $\mathcal{C}$ are invertible, but invertible in a weak sense, because in a Kan complex $1$-simplices are only invertible up to a $2$-simplex witnessing this. Note that a Kan fibrant replacement will have the same vertices still. All of this hopefully makes the following fact believable.
Fact. The underlying homotopy type/A Kan fibrant replacement of $N\mathcal{C}$ is the $\infty$-groupoid that is the $\infty$-categorical localization of $\mathcal{C}$ at all of its morphisms.
Now, if we are given a double category $\mathcal{D}$ (I will use the terminology double category, and not bicategory, since for me the latter is a $2$-category of some sorts), then it also has a nerve $N\mathcal{D}$ whose $(m,n)$-simplices are informally diagrams in $\mathcal{D}$ indexed by the shape $[m]\times[n]$, in which the $[m]$ consists of horizontal morphisms and the $[n]$ of vertical morphisms. The geometric realization of this bisimplicial set can be modelled by first taking the diagonal simplicial set $\mathrm{diag}(N\mathcal{D})$, and then looking at the underlying homotopy type of this simplicial set. The objects of this simplicial set are hence the objects of $\mathcal{D}$, and its $1$-edges are squares in $\mathcal{D}$. We can thus think of $\mathrm{diag}(N\mathcal{D})$ informally as something that tries to be a $1$-category with the objects of $\mathcal{D}$, and in which a morphism is a square in $\mathcal{D}$. As such, we can almost think about a morphism as something of the form $a\xrightarrow{v} b\xrightarrow{h} c$ or $a\xrightarrow{h} b\xrightarrow{v}c$, where $v$ and $h$ are some vertical resp. horizontal morphisms in $\mathcal{D}$. We can declare furthermore that two such compositions are ''homotopic'' if there is a square in $\mathcal{D}$ that has both these composites as its boundary, but a morphism in $\mathrm{diag}(N(\mathcal{AB}))$ truly need the data of the square (the ''homotopy''). By the fact above, the underlying homotopy type of $\mathrm{diag}(N\mathcal{D})$ therefore informally is the $\infty$-categorical localization of this ''trying-to-be category''.
If $\mathcal{B}$ is a category, considered as vertically constant bisimplicial set, then $\mathrm{diag}(N\mathcal{B})$ truly is the simplicial set $N\mathcal{B}$, where this nerve is the good-old $1$-categorical nerve, so we understand what its underlying homotopy type is: it is $\mathcal{B}[\mathrm{all}\,\mathrm{morphisms}^{-1}]_\infty$. On the other hand, we would like to say that the ''trying-to-be category'' $\mathrm{diag}(N(\mathcal{AB}))$ has the same morphisms as $\mathcal{B}$. Being imprecise and ignoring the squares, we can write any morphism in the former as $x\xrightarrow{f} y\xrightarrow{g} z$, where either the first or the second morphism lies in $\mathcal{A}$. Suppose $f\colon x\to y$ lies in $\mathcal{A}$. Then the morphism $x\xrightarrow{id_x} x\xrightarrow{gf} z$ also consists of a morphism in $\mathcal{A}$ followed by a morphism in $\mathcal{C}$, as $x\in\mathcal{A}$ by assumption that $f$ lies in $\mathcal{A}$ (nothing special about $\mathcal{A}$ is needed here), and this morphism is equal to our original morphism since there is a square in $\mathcal{AB}$ witnessing that. Suppose now that $g\colon y\to z$ lies in $\mathcal{A}$. Then the morphism $x\xrightarrow{f} y\xrightarrow{g} z$ is again equal to $x\xrightarrow{id_x} x\xrightarrow{gf} z$, and the latter is again a morphism in $\mathcal{A}$ followed by one in $\mathcal{B}$ precisely because $\mathcal{A}$ is a wide subcategory (i.e. it contains all identity morphsims in $\mathcal{B}$). So, if $\mathcal{A}$ is a wide subcategory of $\mathcal{B}$, then we can write any morphism of $\mathrm{diag}(N(\mathcal{AB}))$ as an identity morphism followed by a morphism in $\mathcal{B}$ (that is hence uniquely determined by the corresponding morphism in $\mathcal{AB}$). Informally, this shows that the ''category'' that $\mathrm{diag}(N(\mathcal{AB}))$ tries to be consists of the objects of $\mathcal{B}$ and the morphisms of $\mathcal{B}$, so essentially $\mathrm{diag}(N(\mathcal{AB}))$ thinks it is $N\mathcal{B}$ already. We need to pass to underlying homotopy types because we definitely do not have an isomorphism of simplicial sets: there are many many more squares in $\mathcal{AB}$ than in the constant double category $\mathcal{B}$, and our $1$-morphisms in the ''category'' corresponding to $\mathrm{diag}(N(\mathcal{AB}))$ (which are squares in $\mathcal{AB}$) that induce the same morphism in $\mathcal{B}$ are in the simplicial set only witnessed to be equivalent by $2$-simplices corresponding to various diagrams in the double category $\mathcal{AB}$. They are not strictly equal in the simplicial set. And then we do get that the underlying homotopy type of $\mathrm{diag}(N(\mathcal{AB}))$ also models $\mathcal{B}[\mathrm{all}\,\mathrm{morphisms}^{-1}]_\infty$ in case $\mathcal{A}$ is a wide subcategory. Moreover, this intuition also tells us why we cannot generally expect the result to also hold if $\mathcal{A}$ is not a wide subcategory.
