Yes, crossover distortion.
The key to understanding why this happens, is to look at the LM358's internal schematic. Refer to figure 16 on page 13 of the TI datasheet, if you can't see my schematic. For the moment, focus on the output stage: Q5, Q5, Q7 and Q13.
http://www.ti.com/lit/ds/symlink/lm358-n.pdfQ5 and Q6 are a Darlington pair. They act as one transistor, with a very high current gain, but double the 0.6V turn on voltage of a single transistor. It works when current is flowing out of the output, i.e. sourcing current.
Q13 sinks current, i.e. it's used when current is flowing into the output pin.
The bases of Q13 and Q5 are connected together.
When the current in the output pin changes direction, i.e. it goes from sourcing to sinking or vice versa, the voltage on the bases of Q13 and Q5 have to change by three diode drops, or about 1.8, which can't happen instantaneously. During this transition point, both Q6 and Q13 will be off, so the output will be unable to supply any current, hence the dead-band in the output. If you could put the oscilloscope probe on the bases of Q5 and Q13 you'd see the voltage jumping by 1.8V, as the waveform crosses zero.
Other things to note about the output stage:
Q7 provides short circuit protection, by shunting the base of Q5 to ground, when the voltage across R
SC exceeds 0.6V. Note that there's no protection against the output being shorted to +V.
The 50µA current sink enables the output to go very close to the negative rail, about 50mV, but it only works when the output stage is sourcing current or sinking well under 50µA. Increase the current sunk by connecting a resistor between the output and +V and the minimum output voltage will shoot up to 0.7V. Using low value feedback resistors can also have this effect, in some standard op-amp configurations. Figuring out which is an exercise left to the reader.

As mentioned above, connecting a resistor between the output and the negative supply will help reduce the crossover distortion, but connecting it the positive supply also works. Try changing the resistor value. Notice how the point where the crossover distortion shifts, depending on the resistor value and whether it's connected to +V or -V? It's because it changes the point when the current flowing though the output pin reverses.