For any positive integer $n$, let $[n] = \{1, 2, \dots, n\}$. Consider a $k$-thin set $A$ with the associated integers $n_1, \dots, n_k$. Let $N$ be the maximum of these integers. We let
$$A + n = \{a + n : a \in A\}.$$
For any fixed integer $L$, let $B = A \cap [L]$. Now $B + n_i$, $i = 1, \dots, k$, are mutually disjoint subsets of $[N+L]$. Also $|B + n_i| = |B|$. Thus $k|B| = \sum |B + n_i| \le L + N$. (Note: this yields $\frac{|A \cap [L]|}{L} \le \frac{1}{k} + \frac{N}{kL}$. This shows that the probability that any positive integer belongs to $A$ is $\le 1/k$.)
Now applying this result to the sets $A_1, \dots, A_m$, with $N$ being the maximum of all the associated integers, we get $|A_i \cap [L]| \le (L+N)/k_i$. Since $\cup_i A_i \cap [L] = [L]$ and letting $p = \sum \frac{1}{k_i}$, we have, by summing over all $i$,
$$L \le (L + N)p, \quad \text{or} \quad (1-p)L \le Np.$$
The above must hold for all positive integer $L$. Since $Np$ is a constant, this is impossible unless $1-p \le 0$ or $p \ge 1$.