Misunderstanding the Taylor Remainder Theorem
You need to specify the interval $I$, the function $f$, the degree $n$, the value of $a$, and (what's most counter-intuitive because of how often we use the symbol), we have to fix a value of $x \in I$. Only after you have specified all of these, the theorem tell you there exists a $c$ between $a$ and $x$ (it may be clearer if you call it $c_x$) such that \begin{align} R_{n,a}(x) = \dfrac{f^{(n+1)}(c_x)}{(n+1)!}(x-a)^{n+1} \end{align}
But of course, everything depends on a pre-chosen value for $x$. If you change $x \in I$, you will have to choose a different value for the $c$.
Edit:
Here's how I'd phrase the theorem (just adding in a few adjectives to make it explicit what is being fixed etc)
Let $I \subset \Bbb{R}$ be a given open interval, let $n \in \Bbb{N}$ be given, and let $f: I \to \Bbb{R}$ be a given $\mathcal{C}^{n+1}$ function. Fix a number $a \in I$; now we denote $P_{n,a,f}$ and $R_{n,a,f}$ to be the $n^{th}$ order Taylor polynomial for $f$ about the point $a$, and the $n^{th}$ order Remainder about the point $a$.
Now, fix a particular number $x \in I$. Then, there exists a number $c$ between $a$ and $x$ such that \begin{align} R_{n,a,f}(x) &= \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. \end{align}
Notice that the number $c$ in the theorem depends on several things: it depends on $f,n,a,x$, but of course, we don't explicitly mention all of these in the notation. It is pretty much only with practice that you'll be able to recognize which quantities depends on which.
Here's another way of phrasing the same theorem:
Let $I \subset \Bbb{R}$ be a given open interval, let $n \in \Bbb{N}$ be given, and let $f: I \to \Bbb{R}$ be a given $\mathcal{C}^{n+1}$ function. Then, for every $a \in I$ (we let $R_{n,a,f}$ mean the $n^{th}$ order Taylor remainder) and any $x \in I$, there exists $c \in I$, lying between $a$ and $x$, such that \begin{align} R_{n,a,f}(x) &= \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. \end{align}
The number of "for all"'s and "there exists" in quick succession may be confusing, but it is very important to recognize which is a bound variable and which is not. I think part of your confusion in the theorem stems from the fact that in the quoted theorem, the author has tried to give a definition of $R_{n,a}$ (namely $R_{n,a} := f - P_{n,a}$) in the same sentence as the actual conclusion of the theorem (which is the final formula for $R_{n,a}(x)$ in terms of $f,n,a,x$ and some number $c$).
Edit 2: Some Additional Remarks
Assuming you have understood my remarks above, let me address your 2nd last paragraph
"Here's where I think I might be messing up. If I consider the interval $(−5,5)$, which is an open interval containing $0$, where $f(x)$ is $(n+1)$-times differentiable, I am unable to come up with a $c$ where the function $e^x$ is identical to $1+x+\dfrac{x^2}{2}+ \dfrac{x^3}{3!}$ in the interval $(−5,5)$. Here is a link to a Desmos page where I tried to find such a $c$."
This is in fact not a coincidence. There is actually no such value of $c$. The proof that there is no single $c$ is actually a very simple proof by contradiction. Let us suppose for simplicity that the interval $I$ is the whole real line $\Bbb{R}$. Suppose, for the sake of contradiction, there exists a $c$, such that for all $x \in \Bbb{R}$ \begin{align} e^x &= \left( 1 + x + \dfrac{x^2}{2!} + \dots + \dfrac{x^n}{n!} \right) + \dfrac{e^c}{(n+1)!}x^{n+1} \quad \text{(for all $x \in \Bbb{R}$)} \tag{$\ddot{\smile}$} \end{align} Notice that the RHS is a polynomial, while the LHS is an exponential, hence can't be a polynomial. This is a contradiction.
If you want to be more explicit about where the contradiction lies, here's one approach: Suppose as a first case, that $n$ is even. Then, the RHS is a polynomial with odd degree; therefore it has a root (this is a simple exercise using the intermediate value theorem). However, the exponential function has no roots. This is a contradiction.
If on the other hand $n$ is odd, then the RHS will be an even degree polynomial. Now, since I wish to stay within the realm of real numbers and not invoke the fundamental theorem of algebra, here's a simple trick: let's integrate both sides of $(\ddot{\smile})$. Then, you'll find that \begin{align} \text{exponential} =\text{polynomial of odd degree} \qquad \text{(everywhere on $\Bbb{R}$)} \end{align} Thus, we're back to case 1. This completes the proof that there is no hope of finding a value of $c$ as you suggested.
Let's break this down a bit. The syntax of the theorem statement actually works like this:
Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$. For all $x$ in $I$, the statement given below is true.
The "statement given below" consists of all the rest of the theorem, including both of the displayed formulas and the text that follows them:
$$f(x) = p_n(x) + R_n(x)$$ where $p_n$ is the $n$th-order Taylor polynomial for $f$ centered at $a$ and the remainder is $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$ for some point $c$ between $x$ and $a$.
In no way shape or form does the theorem say that for all $x$ in $I$, $f(x) = p_n(x) + R_n(x).$ That would be meaningless, because $R_n(x)$ has not even been defined. Instead, for some point $c$ between $x$ and $a$ the two displayed equations are satisfied, and the point $c$ may depend on $x$ (indeed it must depend on $x,$ given the "between" condition), just as $\delta$ may depend on $\epsilon$ in an epsilon-delta proof.
Here is another way the theorem could be stated:
Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$. For all $x$ in $I$, there exists some point $c$ between $x$ and $a$ such that $$f(x) = p_n(x) + R_n(x)$$ where $p_n$ is the $n$th-order Taylor polynomial for $f$ centered at $a$ and the remainder is $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.$$
Or even more directly:
Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$, and let $p_n$ be the $n$th-order Taylor polynomial for $f$ centered at $a$. Then for all $x$ in $I$, there exists some point $c$ between $x$ and $a$ such that $$f(x) = p_n(x) + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.$$
So this is actually the familiar "for all .. there exists ..." that occurs in epsilon-delta definitions as well. The theorem statement in the book may be a little confusing because the "there exists" is worded as "for some" and is tucked away at the very end of the theorem statement instead of just after the "for all".
The way the theorem is stated is fairly typical for calculus textbooks (if my memory serves), and I think the reason is to introduce the notation $R_n(x)$ for the remainder term. It might be a little less misleading if it were written $$R_n(x) = \frac{f^{(n+1)}(c(x))}{(n+1)!}(x-a)^{n+1},$$ writing $c(x)$ rather than just $c$ to remind you that $c$ is not a constant over all $x.$