Alright, let's say we denote the centre of Earth's velocity relative to the sun by the vector $$\vec{v_1}=\left(v_{1_x},v_{1_y},v_{1_z}\right)$$. And the sun has velocity $$\vec{v_2}$$ relative to the centre of the galaxy. And our galaxy has velocity $$\vec{v_3}$$ relative to whatever, and so on... Denote the vector sum of these velocities as $$\vec{v_e}=\vec{v_1}+\vec{v_2}+\vec{v_3}+\ldots$$. So the centre of Earth is moving with velocity $$\vec{v_e}$$ with respect to some reference point somewhere in the universe.

Place an observer at point $$A$$ on the Earth's surface, they will be rotating about Earth's centre with velocity $$\vec{v_o}$$, and thus have a net velocity with respect to our chosen reference point which is simply $$\vec{v_e}+\vec{v_o}$$, where $$\vec{v_o}$$ is tangent (parallel) to the Earth's surface and perpendicular to the radial vector pointing from point $$A$$ to the Earth's centre. Imagine the observer at $$A$$ is hanging from a high bar above the surface and then lets go so they're not touching anything.

Since the observer at $$A$$ isn't touching Earth at the beginning, and we imagine gravity doesn't exist so nothing else is affecting the Earth, the Earth doesn't undergo any acceleration: $$\frac{\Delta\vec{v_e}}{\Delta t}=0$$ ($$t$$ here is obviously time, and $$\Delta$$ refers to changes). So what happens to the observer? We want to calculate the change in the observer's velocity as well: $$\frac{\Delta\left(\vec{v_e}+\vec{v_o}\right)}{ \Delta t }=\frac{\Delta\vec{v_e}}{\Delta t}+\frac{\Delta\vec{v_o}}{\Delta t}=0+\frac{\Delta\vec{v_o}}{\Delta t}=\frac{\Delta\vec{v_o}}{\Delta t}$$.

So we see that any acceleration by the observer can be determined entirely in terms of $$\Delta\vec{v_o}$$, the change in the velocity of the observer with respect to Earth, regardless of how Earth is moving relative to any other reference points. Like the Earth, the observer does not feel any forces from the rest of the universe, nor are they touching Earth at the start, so without gravity, they feel no forces and hence no accelerations either: $$\frac{\Delta\vec{v_o}}{\Delta t}=0$$. Therefore the observer maintains a constant velocity $$\vec{v_o}$$ relative to the centre of the Earth, until such time as they come into contact with it. But since $$\vec{v_o}$$ was tangent to the surface at $$A$$, the observer's starting point, this means the observer never collides with Earth, but rather keeps drifting off into space.

A similar argument shows that if I jump off the Earth's surface and there's no such thing as gravity, I won't come back down no matter how the Earth is moving relative to any other reference points.