Shift property of the Short-term Fourier transform (STFT)
Of course the STFT changes when the signal is shifted. According to the definition we have

$$\text{STFT}\{x(t)\}=X(\tau,\omega)=\int_{-\infty}^{\infty}x(t)w(t-\tau)e^{-j\omega t}dt\tag{1}$$

If we define a shifted signal $y(t)=x(t-T)$, the corresponding STFT becomes

$$\begin{align}\text{STFT}\{y(t)\}=Y(\tau,\omega)&amp;=\int_{-\infty}^{\infty}x(t-T)w(t-\tau)e^{-j\omega t}dt\\&amp;=\int_{-\infty}^{\infty}x(u)w(t-(\tau-T))e^{-j\omega u}e^{-j\omega T}du\\&amp;=e^{-j\omega T}X(\tau - T,\omega)\tag{2}\end{align}$$

From $(2)$ you see that time shifting results in a phase factor and a shift in the STFT. The magnitudes of the STFTs are related by just a time shift, which is of course identical to the time shift of the signal:

$$|\text{STFT}\{x(t-T)\}|=|Y(\tau,\omega)|=|X(\tau-T,\omega)|\tag{3}$$