As long as you add the mass in a way that does not affect its speed, then the orbit is not changed(your star must be fixed as well).
Lets say the planet(mass $M$) is orbiting at a radius $R$, about a star of mass $M_\star$. The orbital velocity is $$v_1=\sqrt{\frac{GM_\star}{R}}$$. Now, in the comments you stated that you added the mass in a way that does not affect its velocity directly. Simce momentum is conserved, the only way to do this is to give the added mass $m$ a velocity $v_1$ as well at the time it reaches the planet. As you can see, when the planet captures the mass, there is no change in angular momentum ($mv_1R+Mv_1R=(m+M)v_1R$). Now, since theres no change in angular momentum, it will orbit at the same angular velocity. If its the same angular velocity, the radius is the same as well. So it stays in stable orbit. One can get this directly from $v_1=\sqrt{\frac{GM_\star}{R}}$ as well.
What if the mass was at rest and it was captured? Well, then by conservation of linear momentum, the velocity would decrease to $v_2$. Since the velocity decreased, it will go into an elliptical orbit. If the velocity had increased, the orbit could be elliptical, but it can be hyperbolic (greater than escape velocity) and leave the system as well. This depends upon the mass ratio.
If the central mass was not fixed, then the masses orbit around the center of mass(barycenter), and the orbital angular velocity is given by$\omega=\sqrt{\frac{G\mu}{R^3}}$ (note that I'm using angular velocity in this case, as the star and planet will have different velocities). $R$ is the distance between the objects, and $\mu=\frac{MM_\star}{M+M_\star}$ is the reduced mass. One can see that a whole variety of things can happen, depending on how you add the small mass, and on the ratio between the three masses. You may want to analyse this yourself (seeing as it's not part of the question and it's a pretty interesting exercise)
You can take the Newtonian expression for the orbital speed as a function of orbital radius and see what radius corresponds to an orbital speed of $c$, but this is not physically relevant because you need to take general relativity into account. This does give you an orbital radius for light, though it is an unstable orbit.
If the mass of your planet is $M$ then the radius of the orbit is:
$$ r = \frac{3GM}{c^2} $$
where $G$ is Newton's constant. The mass of the Earth is about $5.97 \times 10^{24}$ kg, so the radius at which light will orbit works out to be about $13$ mm.
Obviously this is far less than the radius of the Earth, so there is no orbit for light round the Earth. To get light to orbit an object with the mass of the Earth you would have to compress it to a radius of less than $13$ mm. You might think compressing the mass of the Earth this much would form a black hole, and you'd be thinking on the right lines. If $r_M$ is the radius of a black hole with a mass $M$ then the radius of the light orbit is $1.5 r_M$.
So you can only get light to orbit if you have an object that is either a black hole or very close to one, but actually it's even harder than that. The orbit at $1.5r_M$ is unstable, that is the slightest deviation from an exactly circular orbit will cause the light to either fly off into space or spiral down into the object/black hole.
If you're interested in finding out more about this, the light orbit round a black hole is called the photon sphere, and Googling or this will find you lots of articles on the subject.
Best Answer
No, there would be no detectable - and surely no dangerous - changes to the Earth's orbit.
Just for the sake of an argument, imagine that we double our coal reserves by bringing coal from another place - and even some of the precious metals are too expensive to be brought by spaceships at this moment. ;-)
The Earth's coal reserves are something like 1 trillion tons which is $10^{15}$ kg. Let's bring this amount from another celestial body - it's about 10 orders of magnitude less than what we can do now but let's imagine we double our coal reserves in this way.
The Earth's mass is $6\times 10^{24}$ kg, so the coal reserves are approximately $10^{-10}$ of the Earth's mass. Now, if all the extraterrestrial coal landed by a slow speed relatively to the Earth, the Earth's velocity wouldn't change at all; only the mass would increase and the heavier Earth would continue along exactly the same trajectory as before.
But now, imagine that the coal lands at some speed, e.g. the safe speed that the space shuttles used to take. It's about $350$ km/h. If you don't get approximately this low, there's a risk that your coal will burn in the atmosphere.
So if $10^{-10}$ of the Earth's mass has a relative velocity that is $350$ km/h, and let's imagine that all the momentum will go in the same direction - we could make the coal space shuttles land at different places if we wanted - then the velocity of the Earth will change by $350\times 10^{-10}$ km/h which is $3.5\times 10^{-5}$ m/h or $10^{-8}$ m/s. The speed of Earth around the Sun is approximately $3\times 10^{4}$ m/s, so we only change it by a trillionth. Correspondingly, the eccentricity of the Earth's orbit could change at most by one trillionth. We would have a hard time to detect this change.
Obviously, you would need to increase the amount of resources you bring roughly by 10 orders of magnitude (which means by 20 orders of magnitude relatively to what we can achieve today) to produce any threat for the Earth. But even if you brought the whole Moon here to Earth, $7\times 10^{22}$ kg (eight orders of magnitude heavier than the Earth's coal reserves), there would be no significant change of the orbit because the speed of the Moon is approximately the same as the speed of the Earth - they're bound together. Well, the shape of the Earth could change a bit if we tried to incorporate the Moon too quickly. ;-)
You would have to bring a big fraction of Mars to the Earth (Mars is both heavier and has a substantially different speed) to change the eccentricity of the Earth's orbit by a significant amount and I assure you that this will remain in the realm of science fiction for many, many centuries if not forever. If you brought 1/3 of Mars to the Earth, you would also have to build mountains that are 1000 kilometers high - more than 100 times Mount Everest, around the whole Earth. And it would still not be too dangerous as far as the orbital characteristics go. Of course, there could be a danger for the people who suddenly have 1000 kilometers of rock above their heads. ;-)
Your question clearly seems to be an artifact of the unscientific doomsday scenarios that have been presented as science in recent years - e.g. the doomsday caused by CO2 in the atmosphere. Even relatively to the atmosphere, our additions of CO2 are negligible - we're changing the number of molecules in the atmosphere by 2 parts per million (0.0002%) per year. But the atmosphere is just one millionth of the Earth's mass, so our annual CO2 emissions only redistribute something like 2 parts per trillion of the Earth's mass every year. Clearly, all those changes are irrelevant from a "mechanical" viewpoint and they're arguably irrelevant from a climatic viewpoint, too.