We have

$$\log \sum_{k \ge 0} T_k t^k = \sum_{i=1}^n \log \frac{x_i t}{1 - e^{-x_i t}}$$

so if we write

$$\log \frac{x_i t}{1 - e^{-x_i t}} = \log \sum_{k \ge 0} B_k^{+} x_i^k \frac{t^k}{k!} = \sum_{k \ge 1} b_k x_i^k \frac{t^k}{k!}$$

(using the sign conventions explained on <a href="https://en.wikipedia.org/wiki/Bernoulli_number#Notation">Wikipedia</a>) then we straightforwardly have

$$\sum_{k \ge 0} T_k t^k = \exp \left( \sum_{k \ge 1} b_k p_k \frac{t^k}{k!} \right).$$

<a href="https://www.wolframalpha.com/input/?i=taylor+series+log%28x%2F%281+-+e%5E%28-x%29%29%29">WolframAlpha</a> gives that the generating function of the series $b_k$ begins

$$b(t) = \log \frac{t}{1 - e^{-t}} = \frac{t}{2} - \frac{t^2}{24} + \frac{t^4}{2880} - \frac{t^6}{181440} \pm $$

which gives $b_2 = - \frac{1}{12}, b_4 = \frac{1}{120}, b_6 = - \frac{1}{252}$. Plugging the denominators into the OEIS gives <a href="http://oeis.org/A006953">A006953</a>, the sequence of denominators of $\frac{B_{2k}}{2k}$, which suggests the following. We have

$$b'(t) = \frac{d}{dt} \left( \log t - \log (1 - e^{-t}) \right) = \frac{1}{t} - \frac{e^{-t}}{1 - e^{-t}} = \frac{1}{t} \left( 1 - \frac{t}{e^t - 1} \right) =  - \sum_{k \ge 1} B_k^{-} \frac{t^{k-1}}{k!}$$

which gives $b_k = - \frac{B_k^{-}}{k}$ for $k \ge 1$. Altogether we have

$$\boxed{ \sum_{k \ge 0} T_k t^k = \exp \left( \sum_{k \ge 1} - \frac{B_k^{-}}{k} p_k \frac{t^k}{k!} \right) }.$$