Why is the Wavelet Transform not affected by non-stationarity of time series? Here we read:

Wavelet analysis overcomes the problems of non-stationarity in time
series by performing a local time-scale decomposition of the signal,
i.e., the estimation of its spectral characteristics as a function of
time

Why is the Wavelet Transform not affected by non-stationarity of time series?
 A: You can intuitively think of wavelet transforms as a tool that allows you to decompose the signal you want to analyze in different temporal "portions" and inspect and carry out a frequency analysis of that portion of the signals.
For this reason the wavelet transform is not affected by the non-stationarity of the time series, since you break up your signal in portions where frequencies are "approximately constant" and carry out independent analysis on that portion.
Note that there is a trade-off between time and frequency resolution: the Heisenberg uncertainty principle states that increasing resolution in time, corresponds to reducing the resolution in frequency. The time and frequency accuracy are hyperparameters that you choose based on the time-frequency dynamics of the signal you analyze.
A: 
Wavelet analysis overcomes the problems of non-stationarity in time
series by performing a local time-scale decomposition of the signal,
i.e., the estimation of its spectral characteristics as a function of
time.

This claim is at best misleading and, if it were more specific, incorrect.

*

*First, the term "spectral characteristics" only makes sense for (covariance-)stationary series.


*Second, wavelet transform does not "overcome...non-stationarity" any better than, other approaches.
To make the comparison concrete, take the Fourier transform as alternative approach.
Fourier Transform
For stationary series, one can compute the Discrete Time Fourier Transform (DFT) $a(\omega_j)$. This in turn gives periodogram $I(\omega_j) = |a(\omega_j)|^2$, which
captures frequency content of the series at $\omega_j$ in the frequency domain.
Wavelet Transform
Similarly, one can also compute the wavelet coefficients $\psi_{jk}$, using the Discrete Wavelet Transform, or DWT (the non-overlap version, say).
The coefficient $\psi_{jk}$ is the "$k$-th coefficient at $j$-th resolution level".
You can loosely view "high resolution" as analogous to "high frequency".
To be more precise, for a specific wavelet basis, you may compute the corresponding filter in the frequency domain, so that a given level of resolution j roughly corresponds to a subset of the frequency domain.
So $\sum_k | \psi_{jk} |^2$, analogous to $I(\omega_j)$, can be interpreted as the "energy content of the series at resolution $j$.
Non-stationary Series
The frequency domain perspective can be extended in an ad hoc way to non-stationary series for both DFT and DWT (it's ad hoc because point 1 above).
If you generate a non-stationary series and compute its DFT, you will find very large values for the periodogram close to zero frequency. Similarly, the DWT will have large values at low levels of resolution.
These observations have been applied to design unit root tests (statistical tests intended to distinguish stationarity vs non-stationarity), based on the DFT and DWT. Neither approach is uniformly superior to the other. The quoted statement is not correct.
Further Comment
Wavelets do, however, offer advantage for long-range dependent data (e.g. fractional Gaussian noise) relative to other approaches, but that's a different discussion.
