1. Assume the contrary: Suppose there exists a set $A$ such that the error term in the formula is $o(\log\log x)$. Specifically, assume:
$$\sum_{n \le x} \tau_A(n) = x + \varepsilon(x) \log\log x$$
where $\lim_{x \to \infty} \varepsilon(x) = 0$.

2. **Express $\tau_A(n)$**: Recall that $\tau_A(n)$ is the number of divisors of $n$ in $A$. Therefore, we can write:
$$\sum_{n \le x} \tau_A(n) = \sum_{n \le x} \sum_{a \in A, a|n} 1 = \sum_{a \in A} \sum_{n \le x, a|n} 1 = \sum_{a \in A} \left\lfloor \frac{x}{a} \right\rfloor$$

3. **Sum over $A$**: Let $A = \{a_1, a_2, \ldots\}$ be arranged in increasing order. Since $\sum_{i} \frac{1}{a_i} = 1$, we can denote $\sum_{i=1}^n \frac{1}{a_i} = 1 - \varepsilon_n$, where $\varepsilon_n > 0$ is a decreasing sequence converging to 0. Hence, $a_n \approx \frac{1}{\varepsilon_{n-1} - \varepsilon_n}$.

4. Rewrite the sum: We have:
$$\sum_{n \le x} \tau_A(n) = x \sum_{a_i \le x} \frac{1}{a_i} - \sum_{a_i \le x} \left\{ \frac{x}{a_i} \right\}$$
where $\{ \cdot \}$ denotes the fractional part.

5. Approximate the sum: Using the approximation $\sum_{a_i \le x} \frac{1}{a_i} \approx 1 - \varepsilon_n$ for $a_n \le x < a_{n+1}$, we get:
$$\sum_{n \le x} \tau_A(n) = x (1 - \varepsilon_n) - \sum_{a_i \le x} \left\{ \frac{x}{a_i} \right\}$$
Thus:
$$x (1 - \varepsilon_n) - \sum_{a_i \le x} \left\{ \frac{x}{a_i} \right\} = x + \varepsilon(x) \log\log x$$

6. Simplify the equation: Rearrange the terms to get:
$$-\varepsilon_n x - \sum_{a_i \le x} \left\{ \frac{x}{a_i} \right\} = \varepsilon(x) \log\log x$$

7. Analyze the second term: The second term on the left-hand side, $\sum_{a_i \le x} \left\{ \frac{x}{a_i} \right\}$, can be significant. For some choices of large $x$, this term can be approximated by $\sim n$.

8. **Rate of $a_n$**: This implies that the rate of $a_n$ is $e^{e^{n/\varepsilon}}$, where $\varepsilon > 0$ can be made as small as we want. However, such fast convergence would make the first term on the left-hand side of the equation too large, making it impossible to satisfy the equation.

9. Claim: There exists as large $x$ as we want for which:
$$F(x) := \sum_{a_i \le x} \left\{ \frac{x}{a_i} \right\} > c \cdot n \quad \text{for} \quad a_n \le x < a_{n+1}$$
where $c$ is an absolute constant.

10. Proof of the claim: (Sketch) We prove there exists $x \in [a_n, 2a_n]$ for which $\sum_{i=1}^n \left\{ \frac{x}{a_i} \right\} > c \cdot n$ (with $c = 2^{-7}$). This is enough to prove the claim since there are infinitely many $n$'s with $a_{2n} > 2a_n$ (otherwise $\sum \frac{1}{a_i}$ would diverge).

11. **Divide $a_i$'s into groups**: Let $A_k := \{a_i \in [2^{k-1}, 2^k) : 1 \le i \le n \}, k = 1, 2, \ldots, m$. Let $I_0$ be the interval $[a_n, a_n + 2^{m-1}]$. It's easy to show we can find an interval $I_1$ inside $I_0$ with length $|I_0|/4$ such that $\left\{ \frac{x}{a_i} \right\} \geq 1/8$ for at least $|A_m|/8$ terms $a_i$ in $A_m$ and every $x \in I_1$. Hence:
$$\sum_{a \in A_m} \left\{ \frac{x}{a} \right\} \geq \frac{|A_m|}{64}$$
Then we repeat the same argument with $A_{m-2}$ and $I_1$ and so on, getting there exists $x$:
$$\sum_{a \in A'} \left\{ \frac{x}{a} \right\} \geq \frac{|A'|}{64}; \quad A' = A_m \cup A_{m-2} \cup A_{m-4} \cup \ldots$$
The same can be made for $A'' := A_{m-1} \cup A_{m-3} \cup \ldots$ and finally the result follows with $c = 2^{-7}$.

12. Conclusion from the claim: Using the above claim, from the equation we get that for any $\varepsilon > 0$ there exist infinitely many $n$'s such that:
$$a_n > e^{e^{n/\varepsilon}}$$

13. Implication: From this, it easily follows there are infinitely many $n$'s satisfying:
$$a_n > (a_1 \cdot a_2 \cdots a_{n-1})^2$$

14. Estimate the first term: For $x \in [a_{n-1}, a_n)$, the first term in the left-hand side of the equation is estimated as:
$$\varepsilon_n x = \left( 1 - \sum_{i=1}^{n-1} \frac{1}{a_i} \right) x \geq \frac{x}{a_1 \cdot a_2 \cdots a_{n-1}}$$

15. **For $x = a_n - 1$**: We obtain:
$$\varepsilon_n x \geq \sqrt{x}$$

16. Final contradiction: Putting it back in the equation yields:
$$|\varepsilon(x)| \log\log x \geq \sqrt{x}$$
This contradicts with $\lim_{x \to \infty} |\varepsilon(x)| = 0$, since we can find as large $x$ as we want for which the above inequality holds.