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 $B^{t}$.

EDIT: It also holds for $T(t)=\exp(tA)$ and $S(t)=\exp(tB)$. First of all, note that by our norm condition the sets of analytic vectors for $A$ and $B$ are the same, so we can work on this common analytic domain and disregard all the issues coming from unboudedness. If $x$ is analytic then $\|T(t)x\|^2 = \sum_{k=0}^{\infty} \frac{(2t)^{k}}{k!} \langle x, A^{k} x\rangle$ and $\|S(t)x\|^2 = \sum_{k=0}^{\infty} \frac{(2t)^{k}}{k!} \langle x, B^{k} x\rangle$, so the same computation as in the first version gives $\|S(t)x\| \simeq \|T(t)x\|$.

EDIT #2: It turns out that if there exists a constant $C$ such that $\frac{1}{C} \|A^{k} x\| \leqslant \|B^{k} x\| \leqslant C \|A^{k}x\|$ and $D(A^k)=D(B^k)$ for any $k$ then $A=B$. As remarked above, both operators have the same analytics vectors and they form a core, so it suffices to work with them. We can therefore write $\frac{1}{C} \langle x, A^{2k} x\rangle \leqslant \langle x, B^{2k} x\rangle \leqslant C \langle x, A^{2k} \rangle$, i.e. $\frac{1}{C} A^{2k} \leqslant B^{2k} \leqslant C A^{2k}$. Since the function $s \mapsto s^{\frac{1}{2k}}$ is operator monotone, we get $\frac{1}{C^{\frac{1}{2k}}} A \leqslant B \leqslant C^{\frac{1}{2k}} A$, i.e. $\frac{1}{C^{\frac{1}{2k}}}\langle x, A x\rangle \leqslant \langle x, Bx\rangle \leqslant C^{\frac{1}{2k}}\langle x, Ax\rangle$ for any analytic vector $x$. Note that the proof of operator monotonicity of $s \mapsto s^{\frac{1}{2k}}$ follows from an appropriate integral representation of this function, so our analyticity assumption shows that we can perform it in our situation. Letting $k\to \infty$ we get $\langle x, Ax\rangle= \langle x, Bx\rangle$ for any analytic vector $x$. By polarisation and density of analytic vectors we may conclude that $Ax=Bx$ for any analytic vector and therefore $A=B$, since the analytic domain forms a core.