The most famous formula ever is probably $E=mc^2$. We use this fact to express mass in terms of energy.
But what are those energies and masses? You know, $m$ in that formula is not the classical mass, it depends on the velocity. We should write $E(v)=m(v)\cdot c^2$.
In particular, when we set $v=0$, we have the rest mass, $m_0$, to which we associate a rest energy $E_0$. $E_0$ is defined as
$$E_0=m_0\cdot c^2$$
Where $m_0$ is the "usual" mass, the rest mass.
Now, since measuring energies is pretty much easier than measuring masses, we tend to measure rest energies. So we usually know $E_0$. That's why we would write the mass as
$$m_0=\frac{E_0}{c^2}$$
This has an advantage: when $E_0$ is given in multiples of electronvolts, the result is quite beautiful. For example, for an electron, $E_0\simeq 511 keV$, so we can write that the (rest) mass of the electron is $m_0\simeq 511 keV/c^2$.
And of course the $c^2$ is there. If you perform that division, you will get
$m_0\simeq 56,78 \ \frac{keV}{(1 m/s)^2}$; and if you transform $keV$ to Joules, you'll get $9,1\cdot 10^{-31} \frac{J}{(m^2/s^2)}$, which are the same number in kilograms.
So, answering to your first question, yes: the $c$ is actually there.
Now, what really happens is an inconvenient truth: the pysicist are lazy. We don't like writing the same things so many times, so we just relax and don't care haha.
The mass appears in so many equations, many times indeed. For each time the mass appears, there is a factor $E_0/c^2$.
So, what we do is saying: let's stop writing $/c^2$. We do know mass has units of mass, so, whenever we see a mass in $MeV$, we must deduce that it is because there is a hidden $c^2$. It's hidden, but it is there.
So when you read $m_0=0,511 MeV$, you must do a dimensional analysys and say: "hey, units don't match; that means I have to fill the equation with $c$'s until they match".
There's always the same question: where do I add them? I've got different options. Well, both are equivalent. You can fill them multiplying the LHS, or dividing the RHS:
$m_0=0.511 MeV/c^2$, or $m_0c^2=0.511MeV$. They are equivalent.
So the answer is that the $c$ is hidden, omitted for lazyness, but it is actually there. Students hate this because it is really confusing at the beginning.
Personally, I keep hating it, but I can't swim against the stream all the time. I keep writing $c$'s in my notes, and people actually likes it. Everybody prefers seeing all the $c$'s, because reading them is free and helpful, they just don't want to bother to write them, haha.
Natural units
But this is quite little rigurous, so we've invented a good argument: I am using another unit of lenght, let's call it $L$, and I define it as $L=c\cdot(1s)$.
Like this, light travels $3\cdot 10^8$ meters in one second, or $1L$, which is the same.
So I can write that the speed of light is $c=1L/s$. Like this, $c$ has the numerical value of $1$.
This is good, because ¡, liek this, the energy $E_0$ (which is energy), and the mass, expressed in "energy / c² in this new unit" will have the same number: $511 keV$.
The only difference now is that $E_0=511keV$ is a complete datus, and $m=511 keV$ is lacking a division of $c^2$, but that would be dividing by $(1 L/s)^2$, so the only problem now is units, and it is no longer the numerical value.
When we used the international system, $m_0=511 keV/c^2$, and that $c^2$ was completely there.
When we use these "natural units", $m_0=511keV$, and the only difference is that those keV are actually divided by a new unit, but the numebr doesn't change.
Both interpretations are equivalent.
