S-unit equation and small sets of places

I think that the precise statement you ask for is false, but I have a feeling that current bounds on number of solutions to $S$-unit equations are not good enough to prove this, even for $K=\mathbf{Q}$.

What I can prove is that it is false if you require $T_x$ to be the subset of the specified size consisting of the smallest primes of $S_x$. (And it seems unlikely that using larger primes would be much better; this is why I think that the precise statement is false.)

Proof: Let me assume $K=\mathbf{Q}$ for simplicity. Let $f(x)$ be the number of solutions to the $S_x$-unit equation. We have $T_x=S_y$ where $y=x^{1/2+o(1)}$, so if $c$ exists, we would have $f(x^{1/2+o(1)}) \ge c f(x)$ for all $x$. Iterating this shows that $f(x)$ is bounded by a polynomial in $\log x$ as $x \to \infty$. But $f(x) \ge x-1$ because of the solutions $a+(1-a)=1$ for $a=2,\ldots,x$. This is a contradiction.

This is an old question, but maybe it's worth mentioning that a conjecture of Erdös, Stewart, and Tijdeman [1] says that if $P_s=\{2,3,\ldots,p_s\}$ is the set of the first $s$ primes, then the number of solutions to the $S$-unit equation in $\mathbb Q$ should satisfy $$ \exp(s^{2/3-\epsilon}) \le \text{\# of sols using $P_s$} \le \exp(s^{2/3+\epsilon}) \quad \text{for all $s>s_0(\epsilon)$}. $$ If this is true, and if (as Bjorn suggests) the smaller primes are the ones yielding the most solutions, this would show that if one takes the square root of the number of primes, then the number of solutions decreases by a large (and reasonably calculable) amount.

In the same article, they prove that there are sets $S$ containing $s$ primes so that the number of solutions is roughly $$ \exp\left( \sqrt{\frac{s}{\log s}} \right). $$ Their clever construction take a lot of initial primes and throws in some carefully selected larger primes.

[1] Erdös, P.; Stewart, C. L.; Tijdeman, R. Some Diophantine equations with many solutions. Compositio Math. 66 (1988), no. 1, 37–56. MR0937987