The only forces that have a significant impact on the motion of Earth's center-of-mass are gravitational forces. The Earth is "freely falling"; in the language of general relativity, the modern theory of gravity, it is moving along a geodesic.
Because of this fact, it is automatically guaranteed that the spacetime in the vicinity of the Earth's world line is flat to first order; the metric and its first derivatives vanish. (The first derivatives are equivalent to the Christoffel symbol which therefore vanishes at the Earth's center, too.)
This is only modified at the second order: the spacetime curvature (the Riemann tensor) is nonzero near the Earth. Equivalently, the spacetime curvature prevents us from setting the metric tensor equal to the flat spacetime metric at the second order. We may have the metric schematically of the form
$$ g_{\mu\nu} (\vec x) \sim \eta_{\mu \nu} + [R_{\alpha\beta\gamma\delta}] [x^\pi x^\rho] $$
So the metric is flat up to corrections that go like $x^2$ where $x$ is the deviation from the Earth's center. These corrections generically manifest themselves as tidal forces; the greatest contributions come from the Moon and the Sun; other planets may matter, too. The non-inertiality of the Solar System as a whole; and the non-inertiality of our local cluster etc. gives increasingly negligible contributions because the tidal forces decrease with the typical distance scale faster than the force itself.
All other deviations from the flat metric are smaller than that. In other words, the tidal forces are the greatest error that you get if you assume that the Earth is an inertial system floating in an empty space; everything else is smaller.
In the text above, I assumed that you use the non-rotating frame of Earth, one that has a fixed orientation relatively to the stars. You may also use a rotating frame that is fixed relatively to the spinning Earth's surface. The inertiality of this rotating frame is obviously violated by the centrifugal and Coriolis forces (and relativistic corrections to them, including frame-dragging etc.).
The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.
When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 \times10^{-3}$ or $0.3\%$ of their weight.
So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.
This of course has exceptions, as cited in njspeer's answer.
If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.
Best Answer
Your argument is actually more or less right, but some of the details are wrong.
First you have to realize that Newtonian mechanics and general relativity have different definitions of an inertial frame. According to Newtonian mechanics, the coffee cup sitting on my desk right now defines a (very nearly) inertial frame, but a falling rock is extremely noninertial, because the rock has an acceleration of 9.8 m/s2. According to GR, free-fall is the preferred inertial state, so the rock is considered to define an inertial frame, but the coffee cup has a proper acceleration of 9.8 m/s2.
The Newtonian definition is actually impossible to define 100% rigorously, but traditionally the "fixed stars" have been taken as a pretty good standard for Newtonian frames. Any frame in which the stars have a very small acceleration is considered a very good inertial frame.
So if you have the Newtonian definition in mind, then your argument only goes wrong at the end, where you refer to a "very fast speed." What's relevant is the stars' acceleration, not their speed. If a rocket ship is gliding through our solar system at 1,000,000 m/s, then it's an inertial frame. It doesn't matter that the stars have a velocity of -1,000,000 m/s in its frame; what matters is that they have a=0.
According to the Newtonian definition, a frame of reference fixed to a point on the earth's surface is not an inertial frame. You can tell this because in that frame, the stars have large centripetal accelerations. However, the earth-fixed frame comes very close to being inertial, because you can find other frames that are inertial and that differ from it only by a very small acceleration. Therefore experiments on the earth's surface need to be pretty sensitive in order to detect any noninertial effects. The classic example of such an experiment is the Foucault pendulum.
In GR, a frame of reference fixed to a point on the earth's surface is not an inertial frame, and it doesn't even come close to being one. It differs from a valid (free-falling) inertial frame by a huge amount -- an acceleration of 9.8 m/s2. Even an extremely crude experiment can determine this. For instance, I can tell because I feel pressure from my chair on the seat of my pants. A secondary issue is that the earth's frame is rotating, and GR does consider rotating frames to be noninertial as well. (There was a lot of historical confusion on this point, including some early mistakes by Einstein, who thought GR would embody Mach's principle better than it actually did.)