Answer: the sum is equal $1$.
This can be seen as follows:
$$\sum_{k \in A} \frac{1}{k-1} = \sum_{k \in A} \left( \frac{1}{k} \cdot \frac{1}{1-1/k} \right) = \sum_{k \in A} \sum_{i \ge 1} \frac{1}{k^i}$$
Proposition: For $x \in \mathbb{Z}$ the sets $\{(m,n) : x = m^n; m, n \in \mathbb{Z}, m, n \ge 2\}$ and $\{(k,i) : x = k^i; k \in A, i \in \mathbb{N}\}$ have the same number of elements.
Proof of the proposition: Let $x = p_1^{a_1} \dots p_t^{a_t}$ be the prime factorization of $x$. $a = \text{GCD}(a_1, \dots, a_t)$, $a_j = ab_j$ for $0 \le j \le t$ and $y = p_1^{b_1} \dots p_t^{b_t}$. Then $x = m^n$ iff $n \mid a$ and $m = y^{a/n}$. So both of the sets above have $\tau(a) - 1$ elements, where $\tau(a)$ stands for the number of positive divisors of $a$. $\square$

Therefore:
$$\begin{aligned} \sum_{k \in A} \sum_{i \ge 1} \frac{1}{k^i} &= \sum_{m \ge 2} \sum_{n \ge 2} \frac{1}{m^n} \\ &= \sum_{m \ge 2} \frac{1/m^2}{1 - 1/m} = \sum_{m \ge 2} \frac{1}{m(m-1)} \end{aligned}$$
And finally,
$$\sum_{m=2}^{l} \frac{1}{m(m-1)} = \sum_{m=2}^{l} \left( \frac{1}{m-1} - \frac{1}{m} \right) = 1 - \frac{1}{l} \to 1 \text{ for } l \to \infty.$$