I think you'll be unsatisfied with an answer about the gravitational field of the electron--to my knowledge, no one has tested anything involving the gravitational field created by microscopic particles. The closest we've come is tests of gravitational redshift involving scattering microscopic particles in the external field of the Earth.
There, is, however, a known solution to Einstein's equation and Maxwell's equations that represents a black hole with a nonzero charge, known as the Reissner-Nordström metric, given as:
$$ds^{2} = - f dt^{2} + \frac{dr^{2}}{f} + r^{2} d \Omega^{2}$$
Where $f=1-\frac{2M}{r} + \frac{q^{2}}{r^{2}}$. In the context of this metric, there is a difference in the gravitational field induced by the charge from what it would be without the charge--you get the $r^{2}$ variation in what becomes the gravitational potential function for timelike geodesics. This is, in principle, measurable (and a failure to measure it would be a contradiction of either Maxwell's equation or General Relativity, which are both stringently tested--the former moreso than the latter).
One, however, must be careful with what one means by 'active gravitational mass' of a system like this--the ADM Mass of this system is still $M$, and is not modified by $Q$, even if particles near the black hole feel different forces due to the presence of the charged particle.
Finally, as a bit of a interesting aside, it should be noted that the presence of a charge moves the location of the horizon to the location $r_{\pm}=M \pm \sqrt{M^{2}-Q^{2}}$, so there is no horizon at all if $Q>M$. It turns out that if you put the values for the electron mass and charge into this equation, you will find that it predicts that the electron should be a naked singularity.
You ask, "How do we describe mass to the aliens, who don't know about our (g)?" This is an example of a class of questions referred to by Martin Gardner as "Ozma problems." The classic Ozma problem is how we describe to aliens the distinction between right and left; the answer is that we do it by describing the weak nuclear force.
Your statement of your Ozma problem seems a little ambiguous to me. Essentially you're asking how we describe to the aliens a unit of gravitational mass. (You don't say so explicitly, but it seems clear from context that you don't mean inertial mass.) Futhermore, there is a distinction bewteen active gravitational mass (the ability to create spacetime curvature) and passive gravitational mass (what we measure with a balance). Not only that, but your question could be interpreted as asking whether we can compare with the aliens and see whether the value of the gravitational constant $G$ is the same in their region of spacetime as it is in ours.
We can easily establish 1 g as a unit of inertial mass. For example, we can say that it's the inertia of a certain number of carbon-12 atoms.
The equivalence principle holds for us, so presumably it holds in experiments done by the aliens as well. This establishes that our 1 g unit of inertial mass can also be used as a unit for the passive gravitational mass of test particles.
You didn't ask about active gravitational mass, but the equivalence of active and passive gravitational mass is required by conservation of momentum, and has also been verified empirically in Kreuzer 1968. Cf. Will 1976 and Bartlett 1986.
The other issue is whether $G$ is the same for the aliens as for us. Duff 2002 has an explanation of the fact that it is impossible to test whether unitful constants vary between one region of spacetime and another. However, there are various unitless constants that involve $G$, such as the ratio of the mass of the electron to the Planck mass.
A more fundamental difficulty in the fundamental definition of mass is that general relativity doesn't seem to offer any way of defining a conserved, global, scalar measure of mass-energy. See, e.g., MTW, p. 457
Bartlett, Phys. Rev. Lett. 57 (1986) 21.
Duff, 2002, "Comment on time-variation of fundamental constants," http://arxiv.org/abs/hep-th/0208093
Kreuzer, Phys. Rev. 169 (1968) 1007
MTW: Misner, Thorne, and Wheeler, Gravitation, 1973.
Will, “Active mass in relativistic gravity: Theoretical interpretation of
the Kreuzer experiment,” Ap. J. 204 (1976) 234, available online at adsabs.
harvard.edu.
Best Answer
I'm not sure whether these theoretical ideas are is included in what you have in mind. They are only good (and the first , as far as I know, only in theory) for fundamental particles and not for measuring masses of everyday things, but here goes. The second - inference from cross coupling co-efficient between otherwise dispersionless, massless states - is actually the method we use to show that neutrinos have mass, but so far we haven't refined it enough to accurately measure that mass. Still, an inference that the rest mass is nonzero is still highly significant and counts for something IMO. Moreover, we may refine this method to give numbers in the future.
Method 1: Fundamental Particle Dispersion Relationships
This method is to infer the mass of a fundamental particle from experimentally measured dispersion relationships.
A possible fourth quality to add to your list is that mass measures what I call a fundamental particle's "stay-puttability". This is actually the generalisation $E^2 = p^2 c^2 + m_0^2 c^4$ the mass-energy equivalence you cite in disguise. (the equation is simply the pseudo-norm of the momentum 4-vector rewritten).
To look at this idea further, let's think of the Klein-Gordon equation for a lone, first quantised particle, which each spinor component of something fulfilling the Dirac equation must fulfill:
$$\left(-\hbar^2 \partial_t^2 + \hbar^2\,c^2 \nabla^2 - m_0^2\,c^4\right)\psi = 0\tag{1}$$
Hopefully you can pick out $E^2 - p^2 c^2 - m_0^2 c^4=0$ from the unwonted way I've written the equation: recall $i\hbar\partial_t$ is simply the LHS of the general Schödinger equation, so that, by the Schödinger equation, $\hat{H}$ and thus equivalent to the energy observable; also $-i\hbar\nabla$ is the momentum observable. Maxwell's equations can also be thought of as a kind of massless Dirac equation, so that the components of the potential four-vector also fulfill (1) and we can think of the photon as being included in this discussion.
For pure energy eigenstates, $i\hbar\partial_t = \hbar \omega$ and if we Fourier transform (1) into momentum space, we get from (1) the dispersion relationship for the fundamental particle:
$$\omega^2 = k^2\,c^2 +\frac{m_0^2\,c^4}{\hbar^2}\tag{2}$$
so that the group velocity is:
$$v_g = \frac{\mathrm{d}\,\omega}{\mathrm{d}\,k} = \frac{c}{\sqrt{1+\frac{m_0^2\,c^2}{\hbar^2\,k^2}}}\tag{3}$$
Massless particles must always be observed to be travelling at speed $c$, as shown by (3). They are always dispersionless. However, if $m_0$ is nonzero in (3) you can slow a particle down, or "make it stay put" by making the momentum $\hbar\,k$ very small. You can see now from (3) what I mean by mass measures a particle's "stay puttability".
So now you can in theory measure $k$ from matter diffraction experiments, or select for a narrow $k$ from a stream of particles whose mass you are trying to measure using a Bragg grating (for electrons or neutrons, read near-perfect matter crystal). Then you can presumably measure their velocities, within the bounds of the Heisenberg uncertainty principle, by using a matter version of something like a Fizeau-Foucault apparatus: i.e. a sequence of chopper wheels with angular displacements between their slits, so that only particles of a certain velocity, proportional to the chopper wheel angular speed, can make it through the chopper wheels. Then you vary the chopper speed to observe which speeds you detect particles at, and this will let you work out $v_g$. Knowing $v_g$ and $k$ now lets you work out $m_0$ from (3).
Method 2: Cross Coupling Co-efficient Measurement
This, as far as I can understand, is actually the method we use to know that neutrinos have mass. So far it is not very accurate: we can only infer nonzero mass but we haven't refined the method enough to say what that mass is. However, we may do so in the future. The beginning point of this discussion is the Dirac equation for the electron written in a particular way: we write the equations for the so-called Weyl spinors, which are a kind of circular polarisation for the electron:
$$\begin{array}{lcl}\partial\!\!\!/ \psi_L &=& -m\,\psi_R\\\partial\!\!\!/ \psi_R &=& +m\,\psi_L\end{array}\tag{4}$$
Maxwell's equations written in the same form are:
$$\begin{array}{lcl}\partial\!\!\!/ \psi_L &=& 0\\\partial\!\!\!/ \psi_R &=& 0\end{array}\tag{5}$$
That is, on comparing (4) and (5), the electron can be thought of as otherwise two massless, dispersionless particles, mutually tethered together by the cross term $m$; note the two first order equations are uncoupled in the Maxwell equation case. The first massless particle "tries" to zip off at the speed of light. Before this particle gets very far, the cross coupling term $m$ in (4) means that it changes into the other particle, which then also "tries" to zip off at lightspeed, only to be converted back to the first particle and the cycle repeats. This is the phenomenon that Schrödinger called the "Zitterbewegung" (German for quivvering motion) (can you say this word aloud without smiling? - I can't! It's a wonderful example of onomatopoeia). The nett result is that the mutually tethered system - the electron - has a rest mass: confined massless particles always have an inertia, as I discuss in my answer here.
Likewise for the neutrino. It used to be thought that the Weyl equation for the neutrino was the same as (4): three uncoupled, massless Weyl equations for the neutrino flavours. But we experimentally observe that a neutrino shifts between flavours as it propagates. Thus we know that there is a nonzero coupling co-efficient between the flavours, and therefore a mass. So flavour oscillation may in the future be another method for measuring mass.