First, we start with those equations for sine and cosine with an infinite number of terms:
\begin{align}
\sin x &= \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots \right) \\[8pt]
\cos x &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \right) \\
\end{align}
We also take a formula for the exponential function also with an infinite number of terms:
$$ \exp x = e^{x} = \sum_{k = 0}^{\infty} \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
That $\exp$ is the exponential function. Both its domain and co-domain are real so far. Thus, if you feed in a real number to that function, it will spit out another real number.
Those series can be proved independently of the Euler's theorem, without relying on arbitrary-looking definitions.
But, what happens if you feed the function a multiple of $i$? Let's make a function $f$ that does that job:
$$ f(x) = \exp(ix) = e^{ix} $$
You feed in a real number to the function $f$, it gives a non-real number to the $\exp$ function and spits out whatever the $\exp$ gives as an answer. So, this proceeds to:
$$ f(x) = \exp(ix) = e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots $$
Now, considering that $i^2=-1$, that $i^3=-i$ and that $i^4=1$, this can be rewritten as:
$$ f(x) = \exp(ix) = e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots$$
Then, we split it into two sums, one without the $i$s and one with them:
$$ f(x) = \exp(ix) = e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) $$
Replacing those power series with the sine and cosine power series:
$$ f(x) = \exp(ix) = e^{ix} = \cos x + i\sin x$$
Or, simply:
$$e^{ix} = \cos x + i\sin x$$
Finally, what is $f(\pi)$?
$$\begin{align}
f(\pi) &= \exp(i\pi) = e^{i\pi} = \cos \pi + i\sin \pi\\
f(\pi) &= \exp(i\pi) = e^{i\pi} = -1 + 0i\\
f(\pi) &= \exp(i\pi) = e^{i\pi} = -1\\
\end{align}$$
Or simply:
$$e^{i\pi} = -1$$
BTW, this is exactly the power series formula.
About the definition, you could argue that the formula $i = \sqrt{-1}$ is an arbitrary definition. However, the thing that actually is $\sqrt{-1}$ exists independently of what we call it, how we describe it or how we define it. So, someone just gave the name $i$ for it.
Defining something is roughly saying that you give a name to something and a description of this thing. You could define $i$ in many ways, and all of them would result in the same thing. You could define it as "let's call $i$ as the thing that multiplied by itself results in $-1$, even if it does not actually exist". Or, much more simply, you write the definition with $i = \sqrt{-1}$, where the variable to the left of the equals sign is the name that you are giving to something and the expression in the right side is the description of that thing.
However, the case of $e^{ix} = \cos x + i\sin x$ or $e^{i\pi} = -1$ is very different. They are not simple arbitrary definitions and were obtained mathematically from calculations. Defining something is very different from obtaining or inferring something.
You could argue that the equals sign is ambiguous in this case since it doesn't show clearly if the equality is the result of an inference process or simply because someone used it to give a name to something. This is why a few people use $a := b$ or $a \stackrel{\text{def}}{=} b$ for the defining equals, but most people simply don't do that or rely on the textual context to tell the difference or simply don't care at all. Thus, the case of $i = \sqrt{-1}$ is a definition, while $e^{ix} = \cos x + i\sin x$ and $e^{i\pi} = -1$ are the results of inference processes.
Finally, this all shows that those formula can be obtained without needing to resort to integral calculus nor to derivatives nor to using some previously made-up definition of how to extend the exponential function into the $\mathbb{C}$ realm. All you need to have is the power-series formula for sine, cosine and exponential, which can be proved independently beforehand, all in the $\mathbb{R}$ domain. In fact, this formula tells us exactly how to extend the exponential function into the $\mathbb{C}$ realm without having to resort on something made-up. We just fed in a non-real number instead of a real one to the exponential formula and saw if it works and what the heck was the result.