Isolated system: Since the matter, energy, and momentum is fixed, the total number of microstates available that satisfy these constraints is fixed/constant. So is the entropy constant? Yes, if the system is in equilibrium. No, if the system is not in equilibrium. What does that mean in terms of microstates?
If the system is in equilibrium, all these microstates are equally probable and the system visits each of these microstates over the course of time (also known as ergodic hypothesis). Therefore the each micostate has equal probability and $S=k \ln\Omega$. In such a state there is no more increase in entropy possible.
If the system is in non-equilibrium the system doesn't have equal probability of being in every microstate. In fact if the system is stuck in non-equilibrium ( e.g., a hot part and cold part in the box separated by a thermally insulating wall) it cannot access some microstates at all. Hence the system is restricted to fewer microstates. Technically, there is no unique global thermodyanmic state for the whole system and you cannot define entropy. But you can calculate entropy by summing up entropy of different local equilibrium parts (e.g., entropy of the hot part and cold part separately). Since the total number of microstates you can access is small, entropy is lesser than what it could be if you remove that thermally insulating wall letting the system equilibrate. Thus, when you equilibrate more microstates become accessible. Think of the system spreading in the phase space. Thus entropy increases. Once the system has reached complete thermodynamic equilibrium no more entropy increase in possible. All allowable microstates have been made accessible and equally probable!
The bare Boltzmann formula is a bit basic to make sense of for a lone electron. You need to talk about a generalization: one thinks in terms of the Shannon entropy of the state.
We also need to be careful in that the entropy becomes conditional on knowledge that we already have.
An electron in a known pure quantum state has zero entropy: we know its state perfectly and need no information to define it.
If however, if it is in a mixed state, then there is indeed a nonzero entropy. Let's say the spin up / spin down is unknown to us; that is, it is in a classical probabilistic mixture of pure spin states. This situation might have arisen because we have sampled the electron from an ensemble, or it might have arisen through something like the Wigner's Friend thought experiment, where it is known that the is in a spin eigenstate but not which one by Wigner. Wigner's friend has all the measurement results, so right after the measurement, the entropy conditioned on being Wigner's friend is nought. If there are probabilities $p$ and $1-p$ of the electron's being spin up / down respectively, then the entropy conditioned on being Wigner is $-p\,\log p -(1-p)\,\log p$ (multiply by the Boltzmann constant if you want to give it the same dimensions as $Q/T$. More generally, if a quantum system is in a classical mixture of pure states described by Density Matrix $\rho$, then the above formula generalizes to the von Neumann entropy:
$$S=-\mathrm{tr}(\rho\,\log\rho)$$
Now to the temperature of an electron. Temperature is a parameter of a statistical distribution, namely it defines the Boltzmann distribution of an equilibrium ensemble of particles. As such, the notion is not directly applicable to a lone electron: we simply don't have a system of particles in thermodynamic equilibrium! However, we might have sampled the lone electron from an ensemble of electrons in thermodynamic equilibrium at temperature $T$. We can therefore think of its energy state as a classical mixture of energy eigenstates, and the von Neumann entropy of this mixture is (modulo multiplication by the Boltzmann constant) precisely the Gibbs entropy of the ensemble, calculated as a per-particle average entropy. One can then say that the electron is from a population at temperature $T$. If, for example, we add a small amount of heat to the electron population just before the sampling, then the population's entropy changes by $\mathrm{d}Q/T$, and this further entropy, divided per particle, is the change in the entropy attributable to the heat addition of the electron as a mixture of energy eigenstates.
Best Answer
The appropriate mathematical tool to understand this kind of question, and more particularly Dale's and buddy's answers, is large deviation theory. To quote wikipedia, "large deviations theory concerns itself with the exponential decline of the probability measures of certain kinds of extreme or tail events". In this context, "exponential decline" means: probability that decreases exponentially fast with the increase of number of particles.
TL;DR: it can be shown that the probability to observe an evolution path for a system that decreases entropy is non-zero, and it decreases exponentially fast with the number of particles; thanks to a statistical mechanics of "trajectories", based on large deviation theory.
Equilibrium statistics
In equilibrium statistical mechanics, working in the appropriate thermodynamical ensemble, for instance the microcanonical ensemble in this case, one could relate the probability to observe a macrostate $M_N$ for the $N$ particles in the system, to the entropy of the macrostate $S[M_N]$: $\mathbf{P}_{eq}\left(M_N\right)\propto\text{e}^{N\frac{\mathcal{S}[M_N]}{k_{B}}}.$ Naturally, the most probably observed macrostate, is the equilibrium state, the one which maximizes the entropy. And the probability to observe macrostates that are not the equilibrium state decreases exponentially fast as the number of particles goes to infinity, this is why we can see it as a large deviation result, in the large particle numbers limit.
Dynamical fluctuations
Using large deviation theory, we can extend this equilibrium point of view: based on the statistics of the macrostates, to a dynamical perspective based on the statistics of the trajectories. Let me explain.
In your case, you would expect to observe the macrostate of your system $(M_N(t))_{0\leq t\leq T}$, evolving on a time interval $[0,T]$ from an initial configuration $M_N(0)$ with entropy $S_0$ to a final configuration $M_N(T)$ with entropy $S_T$ such as $S_0 \leq S_T$, $S_T$ being the maximal entropy characterizing the equilibrium distribution, and the entropy of the macrostate at a time $t$, $S_t$ being a monotonous increasing function (H-Theorem for the kinetic theory of a dilute gas, for instance).
However, as long as the number of particles is finite (even if it is very large), it is possible to observe different evolutions, particularly if you wait for a very long time, assuming your system is ergodic for instance. By long, I mean large with respect to the number of particles. In particular, it has been recently established that one could formulate a dynamical large deviation result which characterizes the probability of any evolution path for the macrostate of the system (https://arxiv.org/abs/2002.10398). This result allows to evaluate for large but finite number of particles, the probability to observe any evolution path of the macrostate $(M_N(t))_{0\leq t\leq T}$, including evolution paths such as $S_t$, the entropy of the system a time $t$ is non monotonous. This probability will become exponentially small with the number of particles, and the most probable evolution, that increases entropy, will have an exponentially overwhelming probability as the number of particles goes to infinity.
Obviously, for a classical gas, N is very large, such evolution paths that do not increase entropy won't be observed: you would have to wait longer than the age of the universe to observe your system doing this. But one could imagine systems where we use statistical mechanics, where $N$ is large but not enough to "erase" dynamical fluctuations: biological systems, or astrophysical systems for instance, in which it is crucial to quantify fluctuations from the entropic fate.