How does USB superspeed electrical signaling work?
In the example waveforms you show, the voltages are at the pins of the chip, before the capacitive coupling. The transmitter has approximately 0.3V of DC bias because it runs on a positive voltage supply only. If the receiver was not connected, and you instead put a terminating resistor across the pair, you would find the average DC bias (after the capacitors) to be zero.  With the receiver connected, you will find that the receiver is actually adding a DC bias to the signal, that DC bias being close to its transition threshold.