The labelled constructions translate into the usual exponential-generating-function rules

In the labelled product on an $n$-label set, choosing the $k$ labels sent to the $\mathcal{A}$-part contributes $\left(\genfrac{}{}{0px}{}{n}{k}\right)$ possibilities, and then one chooses an $\mathcal{A}$-object on those labels and a $\mathcal{B}$-object on the complement. Thus the size-$n$ coefficient is ${\sum }_{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)a_k{b}_{n-k}$, which is exactly the coefficient rule for the product of exponential generating functions.

For $\mathrm{SEQ}(\mathcal{A})$, a sequence of length $r$ is an $r$-fold labelled product of $\mathcal{A}$ with itself, so its EGF is $A(x{)}^r$. Summing over all $r\ge 0$ gives ${\sum }_{r\ge 0}A(x{)}^r$, and because $A(0)=0$ this formal geometric series equals $1\mathrm{/}(1-A(x))$.

A labelled set of exactly $r$ $\mathcal{A}$-objects is the same data as an ordered $r$-tuple of pairwise disjoint $\mathcal{A}$-objects modulo permutation of the $r$ components. Therefore its EGF is $A(x{)}^r\mathrm{/}r!$, and summing over $r\ge 0$ gives $\exp (A(x))$ by Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws.

A labelled cycle of exactly $r\ge 1$ $\mathcal{A}$-objects has $r$ linear representatives, so its EGF is $A(x{)}^r\mathrm{/}r$. Summing over $r\ge 1$ gives ${\sum }_{r\ge 1}A(x{)}^r\mathrm{/}r=\log (1\mathrm{/}(1-A(x)))$ by Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws.

In a boxed product, the smallest label lies in the $\mathcal{A}$-part. On size-$n$ labels this is equivalent to choosing a pointed $\mathcal{A}$-object on some $k\ge 1$ labels, with the distinguished label forced to be the smallest, and then a $\mathcal{B}$-object on the remaining labels. Pointing contributes the derivative $A^{\mathrm{\prime }}(x)$, so the derivative of the boxed-product EGF is $A^{\mathrm{\prime }}(x)B(x)$. Since no boxed product has size $0$, the constant term is $0$, and integrating from $0$ to $x$ gives the displayed formula.

Steps 1.1-3.1 are exactly the labelled symbolic-method rules claimed in the statement.