The set of tilings (which I'll call $T$) corresponds in number to the points on a line, so this essentially claims that $\lvert T\rvert=\lvert\mathbb R\rvert=\mathfrak c$. Instead of actually establishing a bijection, you can also establish two injections.
Upper bound on the cardinality
$\lvert T\rvert\le\lvert\mathbb R\rvert$ is easy: interpret $L,R,l,r$ as digits $0,1,2,3$ and the whole infinite sequence as a decimal fraction. Since the digit $9$ does not occur, you don't have to worry about things like $0.99\!\ldots=1$ so you know that every infinite sequence corresponds exactly to one real number. So you have an injective map from $T$ to $[0,1)\subset\mathbb R$.
Uncountability of sequences
$\lvert T\rvert\ge\lvert\mathbb R\rvert$ is harder. You could interpret rational numbers as fractions with base $4$, so every digit would correspond to one of the letters. You could also do the following preprocessing:
$$f(x):=\begin{cases}
L,g(x) &\text{if } x\in[0,1) \\
R,g(1/x) &\text{if } x\in[1,\infty) \\
l,g(-x) &\text{if } x\in(-1,0) \\
r,g(-1/x)&\text{if } x\in(\infty,-1]
\end{cases}$$
This will map any $x$ to the range $[0,1]$ which can then be used as input to the second function $g$ that represents that number in base $4$ and thus turns it into a sequence. You could write $1=0.333\!\ldots_4$ or tweak the definition of $f$ a bit (e.g. using $1-g(1/x)$ in the second case) to ensure that you only translate numbers from the range $[0,1)$ so that you can concentrate on the digits after the point.
Uncountability of point labels
So at this point, you have an injective function from $\mathbb R$ to infinite sequences in these four letters. But now you have a technical problem:
Not all possible sequences of the four symbols can be produced this way
You have to work out the exact conditions of which sequences can and which can not be produced in this way. My guess is that you can show that there are only countably many forbidden sequences, so the remaining sequences would still be uncountably many.
Uncountability of tilings
As Daniel reminded me in a comment, what we have worked to so far (at least if you manage to show the countability of the sequences which are not tilings) demonstrates that there are uncountably many sequences describing a pattern seen from a given tile. But there are infinitely many tiles from which you could label the pattern and it would still be the same pattern that should be counted only once.
Two sequences denote the same pattern if they agree after some point $k$. So you can generate alternate labels for the same pattern by subsequently modifying elements of the sequence, starting at the first. You have four options for the first sequence item, and for each of them four for the second, and so on. You could explore all possible sequence modifications in a breadth first search, therefore there are only countably many (i.e. $\lvert\mathbb N\rvert=\aleph_0$) of these.
So why does that not break the uncountability of $T$? I'm not sure how you feel, but (at least at the moment) a counter-argument works best for me. Suppose there were only countably many tilings in $T$, represented by any suitable sequence for each. Then you could use the common $\mathbb N^2$ iteration: Take the original label of the first tiling, then the original label of the second tiling followed by the first modification of the first tiling, then the original label of the thirs tiling followed by the first modification of the second tiling followed by the second modification of the first tiling, and so on. In the end you'd enumerate all point labels, so there could only be $\lvert\mathbb N^2\rvert=\lvert\mathbb N\rvert=\aleph_0$ of these. This is a contradiction to the uncountability of label sequences we showed before.
Now that I think about it, this only shows $\lvert T\rvert>\aleph_0$ but one would need the continuum hypothesis to deduce $\lvert T\rvert=\mathfrak c$ from this. So if anyone has a better explanation (which still works on an intuitive level) of why $\mathfrak c/\aleph_0=\mathfrak c$ feel free to edit this post, comment, or provide your own answer.
Best Answer
So, you can literally write a book about this stuff (exhibit a) and so I will try my best to keep it short and sweet (as much as possible at least). I'll also not mention anything about the physical interpretation of aperiodic patterns and tilings which mathematical and solid state physicists are interested in - namely quasicrystals.
You mentioned that periodic tilings can be characterised by a group action of the space that they tile under which the tiling is invariant - let's stick to euclidean $2$-space, $\mathbb{R}^2$ for this post. You can do this in a few ways, you could either look at just the subgroup of the translation group on $\mathbb{R}^2$ under which the tiling $\mathcal{T}$ is invariant, in which case a cocompact periodic tiling has invariant subgroup isomorphic to $\mathbb{Z}^2$, or you can consider the subgroup of the full euclidean group on $\mathbb{R}^2$ under which $\mathcal{T}$ is invariant, including rotations - this is more subtle and will depend on the specific tiling but we understand this case very well too.
The Tiling Space of $\mathcal{T}$
Now, let's also consider the dual topological theory to this setup. Whenever we have a group $G$ acting on a manifold $M$, we can consider the quotient space $M/G$ and so in the case of the translation invariant subgroup for a periodic tiling, we have the space $\mathbb{R}^2/\mathbb{Z}^2$ which is a $2$-torus $T=S^1\times S^1$ (in general, for $n$-dimensional periodic tilings we get the $n$-torus $(S^1)^n$). We'll call this space the tiling space of our tiling $\mathcal{T}$ and denote it by $\Omega_{\mathcal{T}}$ for an arbitrary tiling.
If we want to consider rotations as well, we need to be slightly more clever so that we don't lose too much information and instead consider the orbifold (stacky) quotient of the symmetry group. I won't spend any more time on the rotational version as it can get messy fairly quickly and would require its own new question in my opinion.
Now, how can we view the torus $T$ as an object which gives us information about our tiling? Well it's quite simple really; a point in the torus corresponds to a placement of the origin of $\mathbb{R}^2$ somewhere into our tiling. That is, if we choose some periodic tiling $\mathcal{T}$, including a prescribed origin, then if we move along one of the group elements in $G$ we should end up back to where we started and so locally, no matter how far away from the origin you look, everything about the tiling looks the same. We've effectively placed a metric on the set of all tilings of the plane by those tiles in $\mathcal{T}$ which says "if a translation by a vector $x$ can get me from the tiling $\mathcal{T}$ to the tiling $\mathcal{T}'$ then we say $d(\mathcal{T},\mathcal{T}')=|x|$, unless of course there is a vector of shorter length. So really we should say $$d(\mathcal{T},\mathcal{T}')=\inf\{|x|:\mathcal{T}+x=\mathcal{T}\}$$ and in fact this gives a well defined metric on the set of all tilings of the plane which contain only those tiles appearing in $\mathcal{T}$ and moreover this space is, by construction, homeomorphic to the tiling space $\Omega_\mathcal{T}$. This can therefore be taken as our definition of the tiling space.
What if $\mathcal{T}$ is aperiodic?
Well, this is the question you originally asked and it is a good question - it's the reason we spent so long defining the tiling space in terms of effectively local information of the tiling - remember for a periodic tiling the origin only needs to know whereabouts in a tile it is in order to determine what point that tiling represents in the tiling space. We're now set up to handle aperiodic tilings with only a small modification of the above metric.
First, let's suppose that $\mathcal{T}$ is a tiling of the plane which is aperiodic, so it has no translations that fix it. We can say this in notation as $\forall x\in\mathbb{R}^2\setminus\{0\},\:\mathcal{T}+x\neq\mathcal{T}$. Let's consider the following set of tilings. We'll say that $\mathcal{T}'$ is locally isomorphic to $\mathcal{T}$ if for all $R>0$, the set of tiles within $R$ distance of the origin in $\mathcal{T}'$, which we will denote by $B_R(\mathcal{T}')$, can be found somewhere in the tiling $\mathcal{T}$. In notation $\mathcal{T}'$ is locally isomorphic to $\mathcal{T}$ if $\forall R>0, \:\exists x\in\mathbb{R}, \:B_R(\mathcal{T}')=B_R(\mathcal{T}+x)$. We'll call the set of all tilings locally isomorphic to $\mathcal{T}$ the tiling space of $\mathcal{T}$ and write it as $\Omega_\mathcal{T}$. Notice I haven't told you the topology of the space yet.
Now, from the motivating material before, we want to put some metric on $\Omega_{\mathcal{T}}$ which in some way satisfies our notion of two tilings being close if they look very similar to each other around the origin out to a large radius. I'll put the metric here but note that it's not pretty - there are equivalent ways to define this metric and you can even forgo defining the metric altogether and just define the topology but none of them are particularly pretty - instead, just try to keep the intuition that two tilings are close if, after a little wiggle, they overlap each other around the origin on some very large patch of tiles. For tilings $\mathcal{T}',\mathcal{T}''\in\Omega_{\mathcal{T}}$ we define the tiling metric $d\colon\Omega_\mathcal{T}\times\Omega_\mathcal{T}\to\mathbb{R}$ by
$$d(\mathcal{T}',\mathcal{T}'')=\inf\left(\{\sqrt{2}\}\cup\{\epsilon\mid\:\exists u,v\: B_{1/\epsilon}(\mathcal{T}'+u)=B_{1/\epsilon}(\mathcal{T}''+v)\mbox{ and }|u|,|v|<\epsilon\}\right)$$ where $B_R(\mathcal{T})$ is defined as before to be the set of tiles within a ball of radius $R$ about the origin. It's a nice exercise to show that $d$ as above really is a metric and a slightly easier exercise to then show that if $\mathcal{T}$ is periodic, $d$ is equivalent to the previous metric defined in the periodic case. In this sense, the above metric is at least a proper generalisation of the periodic case.
Most of the hard work is done now. We have our tiling space $\Omega_\mathcal{T}$ and it seems to represent our intuition of what it means for two tilings to be similar to a 'high degree of accuracy' - there are hundreds of papers by the way on trying to understand the topology of these spaces - they are extremely complicated spaces. For aperiodic tilings which are locally finite (so there are only finitely many patches of tiles of a given radius appearing in $\mathcal{T}$) they are connected but not path connected spaces. They are also compact.
They are a bit like a solenoid as shown below, except they are not homogeneous spaces. Given a regularity condition known as repetitivity which I won't write here (a Penrose tiling is both locally finite and reptitive) a tiling whose tiles are polygons meeting edge to edge has a tiling space which fibers over the $2$-torus $T$ with fiber homeomorphic to a Cantor set. You can think of moving around the torus as translating the tiling by small amounts, and then hopping a small discrete distance from one point in the Cantor fiber to another corresponds to making a different choice of placing a tile far away from the origin, but everything else closer in than that tile is the same.
What about the algebra?
I will leave this section fairly vague as this post is already getting very long. As the astute reader may have guessed, our usual bag of tricks for studying these spaces won't really work - we have an uncountable collection of disjoint path connected components, each component of which is weakly contractible, so singular (co)homology and homotopy groups are all pretty useless in this setting. What we need is an invariant which can somehow see the connected components as opposed to the path connected components - for reasons I won't go in to, it turns out that the right tool is Čech cohomology.
For CW complexes the Čech cohomology is isomorphic to the singular cohomology so as a bonus, for periodic tilings, we can recover our original group $\mathbb{Z}^2$ as the first Čech cohomology group! For aperiodic tilings the Čech cohomology is a pretty good invariant, it gives a fair amount of information about the tiling space, which can in turn be translated into information about the tiling. We're still trying to find a full understanding of what information is really embedded in the Čech cohomology. Probably the biggest bonus of Čech cohomology is that we can actually calculate it for a large class of spaces including the Penrose tiling using various sophisticated techniques. The Čech cohomology of the Penrose tiling space $\Omega_P$ is given by
$$\check{H}^0(\Omega_P) \cong \mathbb{Z}\\ \check{H}^1(\Omega_P) \cong \mathbb{Z}^5\\ \check{H}^2(\Omega_P) \cong \mathbb{Z}^8.$$
There are a few other invariants we can associate to tiling spaces including its $K$-theory, the tiling groupoid, the $\lim^1$ invariant, various notions of dynamical equivalence and shape equivalence. For the reader who wishes to read more of what has been outlined here, I recommend the book by Lorenzo Sadun$^{[1]}$ linked above, and the seminal article of Anderson and Putnam (1998).$^{[2]}$
$[1]$ Lorenzo Sadun (2008) Topology of Tiling Spaces. American Mathematical Society, ISBN-13: 978-0-08218-4727-5. ISBN-10: 0-8218-4727-9
$[2]$ Jared E. Anderson and Ian F. Putnam (1998). Topological invariants for substitution tilings and their associated $C^\ast$-algebras. Ergodic Theory and Dynamical Systems, 18, pp 509-537.