Of course, except in general relativity, where the absolute value of the energy-momentum-stress tensor becomes observable through the gravitational field (described as the curvature of spacetime in GR), in general you can only measure energy differences. So what you call ##E=0## of a system is more or less arbitrary. In classical relativistic point-particle mechanics, it's however very convenient to choose the energy and momentum such as to make up a four-vector in Minkowski space.
You come to this idea when you ask how to generalize the Newtonian equation of motion
$$m \ddot{\vec{x}}=\vec{F}$$
to special relativity. From the point of view of Minkowski space and Lorentz transformations, the Newtonian expression has a very complicated transformation law. It's much simpler to look for a formulation of a relativistic equation of motion that is manifestly covariant under Lorentz transformations, i.e., formulated in terms of four-vectors and tensors.
The first idea is to use a simpler parameter to parametrize the spacetime trajectory than the coordinate time ##t##. For a massive particle, which always moves along a time-like worldline, you can use its proper time. It's the time some ideal clock measures which comoves with the particle. It's related to the space-time trajectory by
$$c^2 \mathrm{d} \tau^2=\eta_{\mu \nu} \mathrm{d}x^{\mu} \mathrm{d} x^{\nu}=\left [c^2-\left (\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \right)^2 \right ] \mathrm{d} t^2.$$
Here ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## is the Minkowski pseudometric. Obviously ##\mathrm{d} \tau## is a Lorentz scalar, i.e., it doesn't change under Lorentz transformations. Indeed, it's the time measured by an observer at rest relative to the particle, i.e., it's a time measured in a preferred frame of reference (which however in general is not an intertial system).
Now you can define the four-velocity which is a Minkowski vector:
$$u^{\mu}=\frac{1}{c} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
Note that
$$u_{\mu} u^{\mu} = \eta_{ \mu \nu} u^{\mu} u^{\nu}=1.$$
From this, using the invariant mass ##m>0## of the particle, we define the energy-momentum vector of the particle
$$p^{\mu}=m c u^{\mu}.$$
To see, how it's related to the kinetic energy and momentum of a particle in the non-relativistic limit, we express this in terms of the usual coordinate time, ##t##
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau}=m \gamma \begin{pmatrix} c \\ \vec{v} \end{pmatrix},$$
where
$$\gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}, \quad \vec{v} = \frac{\mathrm{d} \vec{x}}{\mathrm{d} t}.$$
Now assume that ##|\vec{v}/c| \ll 1## (non-relativistic limit). Then you get
$$p^0=m c \gamma=m c (1-\vec{v}^2/c^2)^{-1/2}=m c \left (1+\frac{1}{2} \frac{\vec{v}^2}{c^2}\right) = \frac{1}{c} \left (m c^2 + \frac{m}{2} \vec{v}^2 \right ).$$
Now set
$$E=c p^0=m c^2 + \frac{m}{2} \vec{v}^2.$$
This is, up to the additive constant ##m c^2## just the non-relativistic kinetic energy. As we've seen, this limit occurs, because we wanted a four-vector description of energy and momentum, and that's why it's convenient to include the "rest energy" ##E_0=m c^2## in the energy of the particle.
We also immediately
$$p_{\mu} p^{\mu}=m^2 c^2 \; \Rightarrow \; \frac{E^2}{c^2}-\vec{p}^2=m^2 c^2$$
or
$$E=c \sqrt{m^2 c^2+\vec{p}^2}.$$
A good guess for an equation of motion, generalizing Newton's 2nd Law to a relativistic law now is
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu},$$
where ##K^{\mu}## is the Minkowski four-force. However, not all four components are independent of each other, because from
$$p_{\mu} p^{\mu}=m^2 c^2=\text{const} \; \Rightarrow \; p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=p_{\mu} K^{\mu}=0.$$
You find more on relativistic mechanics of classical point particles in my special-relativity FAQ article:
http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf