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The condition $\|A^k x\| \simeq \|B^k x\|$ can be rewritten, by squaring both sides, as $\langle x, A^{2k} x \rangle \simeq \langle x, B^{2k} x \rangle$, i.e. $A^{2k} \leqslant C B^{2k}$ and $B^{2k} \leqslant C A^{2k}$. Recall now that the function $s \mapsto s^{\alpha}$ is operator monotone for $\alpha \in (0,1]$. It means that if you fix $t$, then you should take any $k \geqslant t$ and consider the function $s \mapsto s^{\frac{t}{k}}$, to conclude that $A^{2t} \leqslant C^{\frac{t}{k}} B^{2t}$ and $ B^{2t} \leqslant C^{\frac{t}{k}} A^{2t}$, from which it follows that $\|T(t)x\|\simeq \|S(t) x\|$. In this answer I assumed that in this setting by semigroups $T(t)$ and $S(t)$ you meant $A^{t}$ and $B^{t}$.