Gravitational Potential is a scalar quantity so can be added algebraically directly for both(or more) bodies.
Also GPE is just Gravitational potential times mass. $$E=\underbrace{\big(\sum P\big)}_{\text{due to all bodies in vicinity}}\times m$$
Now , rest of your aproach is allright ! Continue using this.
Here's an example of why this is done using a gravitational context.
Kinetic energy
The kinetic energy of a system of particles is defined to be $K_\text{tot} = \sum_i K_i$, where the sum runs over all particles in the system. This is just a definition, and choosing to use such a definition allows one to speak of the kinetic energy $K_i$ of the $i$th particle, for example, since it's just one of the terms in a sum.
Okay, that was boring.
Potential energy
Gravitational potential energy $U_\text{g} = -\frac{Gm_im_j}{d_{i,j}}$ is defined for pairs of particles.$^\text{1}$ Mathematically you can see that in the subscripts of the masses $m_i$ and the distance between the two particles $d_{i,j}$. Why? Because to determine potential energy from scratch, you must consider how two objects interact, which in classical mechanics is where forces come in. And forces themselves come in pairs.
Now, the fact that they come in pairs means one must be careful when determining the potential energy in some given situation, otherwise you might over-count. Suppose, for example, there are 4 particles with masses $m_i$, and the distance between the $i$th and $j$th objects is $d_\text{i,j}$. What, then, is the potential energy of the masses and distances are all given? The answer is
$$U_\text{g}=-G\left( \frac{m_1m_2}{d_{1,2}} + \frac{m_1m_3}{d_{1,3}} + \frac{m_1m_4}{d_{1,4}} + \frac{m_2m_3}{d_{2,3}} + \frac{m_2m_4}{d_{2,4}} + \frac{m_3m_4}{d_{3,4}} \right)$$
First, note that since there are six terms you can't make a one-to-one associating of particles with potential energy. Second, if someone tried to associate potential energy with a single particle, then that person might mistakenly include too many terms in their sum:
$$U_\text{g}\ne-G\left( \underbrace{\frac{m_1m_2}{d_{1,2}}}_\text{a term for particle 1} + \underbrace{\frac{m_2m_1}{d_{2,1}}}_\text{a term for particle 2} + \cdots \right)$$
This incorrect expression would over-count the number of terms in $U_\text{g}$.
To avoid such over-counting, and to make our language more inline with the mathematics, we tend to speak of the potential energy of the system.
$^\text{1}$The expression $-\frac{Gm_1m_2}{d_{1,2}}$ is a more general form of the more-common $U=mgh$, but that's a separate issue.
Best Answer
There are many fundamental concepts that both you and Trimok have misunderstood.
First of all, you can completely ignore the mass energy. The kinetic energy in no way "compensates" for the loss in mass energy because no mass energy is lost in the example that you gave. Sure, the mass of the rocket decreased as it ejected propellant to move forward, but that mass hasn't disappeared. The propellant is still out there, floating around in space. You can't just ignore its mass energy because it's not in the rocket anymore, it still exists. In other words your total $E_m$ term is always the same throughout this situation, at no point is matter converted into another form of energy, like kinetic energy as you seem to suggest.
Second, this is conceptually wrong :
You can't compare the inertia of a system with a force (they're not even measured in the same unit). One can't be "greater" than another. Regardless of the mass of an object, however small a force you apply on it it will still have an impact. There's no cutoff. This doesn't matter much because indeed, you can get to a point where this is small enough that you consider yourself to be in a situation where the potentially energy is now maximal (you seem to think that because this is 0 the potential energy has now disappeared, but this is not the case, it is sometimes chosen to be 0 at infinity but keep in mind that this is still an increase because in such conventions this potential energy is negative at lift off). Potential energy increases as you get further from the earth. This is because, intuitively enough, now that you are further from the earth you now have the potential to gain more kinetic energy by falling back towards the earth through a longer distance, gaining more speed in the process.
Finally, let's work out the actual energy distribution throughout all this. Like I said we can ignore mass energy because the total mass is conserved. Instead what we actually need to consider is :
$E_k$ : the kinetic energy of the rocket.
$E_c$ : the chemical potential energy stored in the propellant. It is this energy that will propel the rocket, not mass energy as you seem to think.
$E_g$ : the gravitational potential energy.
In the beginning, the rocket is sitting on the surface of the earth, and $E_c$ is maximal, $E_g$ is minimal, and $E_k$ is just $0$. As you start burning propellant, you release the energy stored in your propellant and $E_c$ begins to decrease as $E_k$ increases. To completely compute $E_k$ you actually have to take into account both the kinetic energy of the rocket and the kinetic energy of the ejected propellant. If you do, you will notice that this is actually not enough to make up for the loss in $E_c$. Indeed, the difference between the two corresponds to the gain in $E_g$ as the rocket goes further and further away from the earth, and indeed we will always have conservation of $E_k + E_c + E_g$.