I've heard you can't or it would be very hard if even possible to accelerate electrons in the absence of a vacuum.

Isn't lightning essentially a stream of electron? Lighting is not in a vacuum.

 

Yes and no. Lightning is a channel of plasma created when the electric field between the earth and the sky reaches what's called the "dielectric breakdown point" - the point at which it becomes conductive. The breakdown point for dry air at room temperature and 1 atm of pressure is roughly 30 kV/cm (or 3 MV/m). When the field gets strong enough, it transports charge carriers out of the valence bands in the molecules in the air into the conduction bands, and charge is free to flow, carrying current (making it act like a highly-resistive conductor).

However, as is the case with any conductor, the mean-free-path (the longest distance a charge can move in a conductor in a straight line) is not terribly long, so the "drift velocity" of a charge in a conductor is very small - cm/s, if even that. They're very fast on that free path (a measure of this in a material is the "charge mobility"), but they bounce around so much in not-so-straight lines that their average velocity is very slow. So imagine how slow it is in air - a highly-resistive material, even when it's conducting.

When lightning forms, then, the electric field strips the electrons out of their molecules' valence bands, ionizing the molecules (which is why the lightning is a channel of plasma). These electrons are free to flow, moving very quickly in very, very short paths, so they don't actually go very far. Much energy is lost as heat and light, which is why lightning plasma has temperatures on the order of thousands of kelvin.

Let's do some math to illustrate this point.
Let's assume we have a bolt of lightning that forms between a very low-to-the-ground thunderhead, and 100% of the energy in the field is going to accelerating the electrons. Cumulonimbus clouds can vary in base height of 500 m to 16 km, so the minimum height of 500 m is what we'll use. Ignoring the fact that the entire cloud also tends to contain part of the lightning strike, and the air between the cloud and ground is most likely not "ideal conditions," the electric field between the ground will exceed 3 MV/m * 500 m = 1500 MV (1.5 GV). A charge trapped in that field has the corresponding energy of 1.5 GeV (1 eV = energy in one electron in a field potential of 1 V). So, naively, the velocity of the electron will be:

\[ v = \sqrt{\frac{2E}{m_e}} = \sqrt{\frac{qV}{m_e}} = \sqrt{\frac{1.6\times10^{-19}\cdot 1.5\times 10^9}{9.11\times 10^{-31}}} = 162 \times 10^8 \]

A quick check realizes this is incorrect: that's 54c, which is very wrong. So relativistic electron speeds would be:

\[ E = \sqrt{p^2c^2 + m_e^2c^4} \]
\[p = \frac{m_ev}{\sqrt{1 - \frac{v^2}{c^2}}}\]
\[\rightarrow E = qV = \frac{m_ec^3}{c^2-v^2}\]
\[\rightarrow v = \frac{c\sqrt{q^2V^2-m_e^2c^4}}{qV}\]

Doing the math gets us velocities on the order of 99.99999423% the speed of light (so, really fast - \(2.997924407\times 10^8\) m/s, to be precise). However, doing some digging reveals that the mean free path in air is about \(7\times 10^{-7}\) m, which is really, really short. It'd take the electron (our time - again, relativistic speeds!) barely any time to bounce that distance (0.00233 ps, or 2.33 as), and in the 30 microsecond time span that a single bolt lasts it'd bounce around 12.8 billion times, but mostly in random directions (biased along the direction of the field, though).

Here's the important part, though: All this also ignores the thermal losses and the slowdown of the charge as it moves around. Lightning averages about 30000 K, so there's a not-insignificant amount of energy loss in the plasma channel itself (thermal energy: \(E = k_BT\), where \(k_B\) is the Boltzmann constant and \(T\) is temperature in kelvin). With this increased temperature, the kinetic energy of the molecules will increase significantly and thus make the electron drift even more "chaotic" and random, reducing the overall drift velocity/mean-free-path.

All this math also ignores the amount of energy required to strip the charges from their respective molecules (it takes 13.6 eV just to ionize a single hydrogen atom, so imagine the amount of energy required to ionize a channel, sometimes several centimeters thick, hundreds if not thousands of meters long of water vapor, nitrogen, hydrogen, carbon-based molecules, and more). So the actual electron velocities would be much lower, and the mean-free-path probably even shorter. This is what frequently contributes to the branches off of lightning; rare is it to see a perfectly straight, branch-free bolt of lightning.

Additionally, energy would be re-radiated off as light in a wide variety of bands (visible, obviously, but also other bands as well - depends on how much the electrons are accelerated and in what ways they are), causing a deceleration of the charges. If you have an appropriate antenna, you can actually pick up a lightning strike without actually seeing it or hearing the thunder, due to the incredible amounts of low-frequency RF noise it creates (which, actually, if you knew what frequencies it generated, you could actually use that to approximate the drift velocities of the lightning, due to treating the distances as wavelengths and going from there).

So, tl;dr: Yes, lightning can accelerate electrons, and it's not generally impossible to accelerate electrons in a non-vacuum, but it's also hard to do so, as you need to overcome the dielectric breakdown of the medium, so you'd need a large field and an enormous amount of energy. Doing so is a terribly inefficient and unguided/uncontrolled way to do it and any potential benefits resulting from accelerated electrons are going to be lost. This also won't just accelerate electrons, but charged molecules as well, creating a plasma, generating heat and adding disorder to the molecular ensemble, and causing a lot of other things to occur, reducing the effective acceleration.

 

https://www.physicsclassroom.com/class/estatics/Lesson-4/Lightning

As the static charge buildup in a storm cloud increases, the electric field surrounding the cloud becomes stronger. Normally, the air surrounding a cloud would be a good enough insulator to prevent a discharge of electrons to Earth. Yet, the strong electric fields surrounding a cloud are capable of ionizing the surrounding air and making it more conductive. The ionization involves the shredding of electrons from the outer shells of gas molecules. The gas molecules that compose air are thus turned into a soup of positive ions and free electrons. The insulating air is transformed into a conductive plasma. The ability of a storm cloud's electric fields to transform air into a conductor makes charge transfer (in the form of a lightning bolt) from the cloud to the ground (or even to other clouds) possible.

A lightning bolt begins with the development of a step leader. Excess electrons on the bottom of the cloud begin a journey through the conducting air to the ground at speeds up to 60 miles per second. These electrons follow zigzag paths towards the ground, branching at various locations. The variables that affect the details of the actual pathway are not well known. It is believed that the presence of impurities or dust particles in various parts of the air might create regions between clouds and earth that are more conductive than other regions. As the step leader grows, it might be illuminated by the purplish glow that is characteristic of ionized air molecules. Nonetheless, the step leader is not the actual lightning strike; it merely provides the roadway between cloud and Earth along which the lightning bolt will eventually travel.

If you generate a plasma (positive ions and free electrons) you can have an electron current (flow means accelerated electrons).
https://arxiv.org/pdf/1404.0509.pdf
A Short Introduction to Plasma Physics

In plasmas powered electrically (i.e. in any electric discharge), the electrons gain energy more
easily from the external electric field than the inert ions, as the electrons are much lighter and thus
electronic currents are much larger.