With $d\mu = \sum a_n\delta_{\lambda_n}$, your question can be rephrased as: Does $\int t^k\, d\mu(t) = 0$ imply that $\mu=0$? Since there are indeterminate moment problems (that is, collections of moments that do not come form one unique measure), it's now clear that the answer is no.

To make this more concrete, you can start out with any limit circle Jacobi matrix; any two of its spectral measures $\nu_1,\nu_2$ will have the same moments, and you can take $\mu=\nu_1-\nu_2$. These measures will automatically have the extra properties you require here (pure point, with discrete disjoint unbounded above spectra) if you take two spectral measures corresponding to two distinct boundary conditions at infinity.

With $d\mu = \sum a_n\delta_{\lambda_n}$, your question can be rephrased as: Does $\int t^k\, d\mu(t) = 0$ imply that $\mu=0$? Since there are indeterminate moment problems (that is, collections of moments that do not come from one unique measure), it's now clear that the answer is no.

To make this more concrete, you can start out with any limit circle Jacobi matrix; any two of its spectral measures $\nu_1,\nu_2$ will have the same moments, and you can take $\mu=\nu_1-\nu_2$. These measures will automatically have most of the extra properties you require here (pure point, with discrete disjoint spectra) if you take two spectral measures corresponding to two distinct boundary conditions at infinity. What is not automatic is that these measures will be supported by $[0,\infty)$, but you can get that, too, since there are indeterminate Stieltjes moment problems.