Most traditional introductions to quantum electrodynamics (QED) don't mention this, but QED is perfectly well-defined mathematically — just not in continuous spacetime. They don't mention it because discretizing spacetime is artificial and messy, and they want to portray QED as something natural and beautiful. They sustain that illusion by using a mathematically ill-defined formulation. When the ill-defined foundation leads to ill-defined results, they use ad-hoc bandages to cover the wounds. That might be good enough for some students, but I'm glad you're not one of those students.
Here's a better perspective on QED. We know it's not a theory of everything. Like most of the theories we use, QED is only an approximation to something more complete. But unlike some of the other approximate theories we use, we don't know how to formulate QED by itself in any pleasant way, at least not if we want the formulation to be mathematically unambiguous.$^*$ The only way we currently know how to do it is by discretizing spacetime. That's unpleasant, but it has one big benefit: answers to questions like the one you're asking here become conceptually straightforward. It's also compatible with experiment, because we can make the discretization scale much finer than any scale where we would expect QED to become a bad approximation anyway, like the electroweak scale.
$^*$ A technical note: Mathematically unambiguous implies nonperturbative, because the small-coupling expansion doesn't converge.
Relationship between bare charge and elementary charge
When we define QED by discretizing spacetime, the so-called bare parameters are perfectly finite real numbers. They are the theory's inputs. We adjust their values until all of the theory's predictions line up with the experimental data as closely as possible — not perfectly, of course, because QED is an approximate/incomplete theory. One of those inputs is the one we call the bare charge $e_0$, which is usually encoded in the quantity
$$
\alpha_0\equiv \frac{e_0^2}{4\pi}
\tag{1}
$$
in natural units. Again, this is a finite real number. The question is, how is this parameter related to the fine structure constant $\alpha=e^2/4\pi$, where $e$ is the elementary charge that we measure in experiments? The relationship between them can be worked out by a standard calculation, with the result (from reference 1)
$$
\alpha = \alpha_0 - \alpha_0^2 \kappa\log(\Lambda/m) + O(\alpha_0^3)
\tag{2}
$$
where $m$ is the electron's mass, $1/\Lambda$ is the scale at which spacetime is discretized, and $\kappa$ is a numerical factor of order $1$. The quantity $\Lambda$ is called a cutoff.$^\dagger$ Most textbooks use an ad-hoc short-cut to derive (2) so that they don't need to deal with the messy discrete-spacetime formulation, but we can also derive it using the discrete-spacetime formulation, and then everything in the calculation is finite from beginning to end.
$^\dagger$ Another technical note: The cutoff theory is gauge-invariant, because the kind of cutoff I'm using here comes from formulating QED as a lattice gauge theory.
To ensure that the theory's predictions match the experimental data, the cutoff $\Lambda$ should be very large compared to $m$. In other words, the discretization scale should be very fine, so fine that experiments can't tell the difference between continuous spacetime and this very-finely-discretized spacetime. The ratio $\Lambda/m$ should be $\gg 1$, but it doesn't need to be infinite, contrary to what some introductions might say. We can keep it finite.
To put this in perspective, recall that $\alpha\approx 1/137$. Suppose we take $\Lambda/m\sim 10^6$. Then equation (2) becomes
$$
\frac{1}{137} \approx \alpha_0 - 14 \alpha_0^2 \kappa + O(\alpha_0^3)
\tag{3}
$$
because $\log(10^6)\approx 14$. Since $\kappa$ is of order $1$, this shows that the input $\alpha_0$ is only a little larger than $1/137$, even though $\Lambda$ is enormously larger than $m$.
The cutoff is obviously artificial, so QED's predictions at experimentally accessible resolutions shouldn't depend on the value of the cutoff. At sufficiently low resolution, any change in the chosen value of $\Lambda$ can be compensated by adjusting the bare parameters — just two of them, in the simplest version of QED. This is one aspect of renormalization. We only need to use one value of $\Lambda$ for all applications of QED, but since the cutoff is artificial, knowing that its precise value doesn't matter is reassuring.
Effective charge
For many applications of QED, expanding in powers of $\alpha_0$ up to some low order is a good enough approximation. We can make the approximation slightly better by re-arranging the expansion so that the expansion parameter is (reference 2)
$$
\alpha(\mu) = \alpha + \alpha^2 \kappa\log(\mu/m) + O(\alpha^3)
\tag{4}
$$
instead, with the effective charge $e(\mu)$ defined by $\alpha(\mu)=e^2(\mu)/4\pi$. This is another aspect of renormalization. By choosing the scale $\mu$ to be close to the energy-scale of the experiment of interest, expanding in powers of $\alpha(\mu)$ gives a better approximation than expanding in powers of $\alpha_0$ or $\alpha$. The higher the energy scale $\mu$ of the experiment, the larger $\alpha(\mu)$ becomes. It never reaches $\alpha_0$ simply because we always take $\Lambda$ to be much higher than any energy scale where QED is ever used, so $\mu$ is never anywhere near that scale. We call $e(\mu)$ the effective charge because we can define it by analogy to how we define the usual elementary charge $e$, but using processes at the specific energy scale $\mu$.
References:
Page 219 in Montvay and Münster, Quantum Fields on a Lattice (Cambridge University Press)
Page 255 in Peskin and Schroeder, An Introduction to Quantum Field Theory (Addison-Wesley)
