Massive cancellations

Let $a_n$ be an increasing sequence of positive integers which grows really fast, say $a_{n+1} > \exp(a_n)$. Take $A = \{10^{-1}, \sum 10^{-a_n}\}$. Then $d_A(a_n) \leq 2\cdot 10^{- a_{n+1}} \leq 2\cdot 10^{-\exp a_n}$, so $A$ cannot be tame.

EDIT. One could replace $1/10$ by some transcendental $0<x<1$ such that $\sum x^{-a_n}$ is transcendental as well. This gives an example of a non-tame $A$ consisting of transcendental elements.

Here's an argument that for $A$ a finite set of algebraic numbers, $d_A(n)$ decays at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$, there's a product formula $$ \prod_v |x|_v=1, $$ where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a suitably normalized absolute value. In order for some absolute value to be small, the product of the others must be large.

For fixed $A$, there is a finite set of places $P$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in P$ ($P$ consists of all the Archimedean places, and the non-Archimedean places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$.

In our case, $K$ is embedded in the real numbers. Let $v_0$ be the valuation coming from this embedding. Since the number of $v$ for which $|x|_v>1$ is bounded, we have \begin{gather} d_A(n)=\min \big\{|x|_{v_0}:x\in S_A(n)\backslash\{0\}\big\}=\min\left\{\prod_{v\neq v_0}|x|_v^{-1}\right\} \geq\min\left\{\prod_{v\in P\backslash\{v_0\}} |x|_v^{-1}\right\}. \end{gather} The right hand side decays at most exponentially, so the left hand side does too.

You can do this without much number theory. View your field as a finite-dimensional vector space over $\mathbb Q$. Then every element acts linearly on the field, so it acts as some matrix with rational entries. The actual element you want to bound is an eigenvalue of this matrix.

We can lower bound it by lower bounding the determinant and upper bounding the other eigenvalues. Observe:

The entries grow at most exponentially, so the other eigenvalues grow at most exponentially. Because the number field is a field, the element is invertible, so the determinant is nonzero. The denominators of the entries grow at most exponentially, so the denominator of the determinant grows at most exponentially.

Then you get a lower bound on one eigenvalue by division and the fact that the numerator of the determinant must be at least $1$.

This is connected to the valuations proof as follows: the valuations at $\infty$ are just the absolute value of the eigenvalues, and the others are related to the powers of different primes dividing (denominators of ) entries of the matrix.