While I agree that, for a prime base $b$, the 2nd identity in the OP can be viewed as a rephrasing of Legendre's theorem on the $b$-adic valuation of $n!$, which in turn is a straightforward consequence of De Polignac's formula (en.wikipedia.org/wiki/De_Polignac%27s_formula), I don't see how this can imply the identity in the general case, which is the one for which I'd like to get a reference. However, thanks for the effort!

[...] (multiplicatively written) grp for which there exist $x,y\in G$ s.t. $x^3=1$ and $y^nx\ne xy^m$ for every $m,n\in\mathbf{Z}$ with $m^2+n^2\ne 0$, e.g., a 2-generator 1-relator grp of the form $\langle u,v \mid u^3=1\rangle$. For $n\in{\bf N}$ we let $Y_n := \{y^k: k=0,\ldots,n\}$. Then, we fix $m,n\in{\bf N}$ and let $A:=Y_mx$ and $B:=xY_n$, so that $x\in A\cap B$. We have $$AB=\{y^hx^2y^k: h=0,\ldots,m, k=0,\ldots,n\}$$ and $$AxB=Y_mx^3Y_n=Y_mY_n=Y_{m+n}.$$ Thus, $|AB|=(m+1)(n+1)$ and $|AxB|=m+n+1$, since $y^hx^2y^k=y^rx^2y^s$ for some $h,k,r,s\in \bf Z$ only if $h=r$ and $k=s$.

I'm accepting this as an answer (I had completely forgot about the thread), but let me remark that the hand-written note on page 250, line 3 (claiming that the inclusion $A+d+B\subseteq C^\ast$ is "trivially impossible in view of $|A+d+B|=|A+B|=|C|>|C^\ast|$", where $A$ and $B$ are finite non-empty subsets of the base set, $G$, of a fixed cancellative monoid $\mathbb G$, $C$ is the sumset $A+B$, and $d\in A\cap B$) is incorrect. The point is basically that $\mathbb G$ is not abelian, even if Kemperman writes it additively. For a counterexample, let $\mathbb{G}$ be a [...]

Cool! This answers Q2 (and hence Q1) in the unbalanced case ($\mu = \infty$ can now be treated in a similar way). But what about the balanced case?