Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not possible. I see why this wouldn't be possible, I'm not sure if it really is the cause which makes that equation illegal though.
But now, apparently, I could do this:
$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x=\lim_{x\to -\infty}\frac{[\sqrt{x^2+5x+3}+x][\sqrt{x^2+5x+3}-x]}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{x^2+5x+3-x^2}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5x+3}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5+3/x}{\sqrt{1+5/x+3/x^2}-1}=-5/2$
which gives me the correct result.
But in the 3th step I used $x^2-x^2=0$, how is that legal?
Also, in the 2nd step I implicitly used:
$-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$
Which also seems to be fine, but why? 

The first step is illegal, not the second. This is because $\sqrt{x^2 + 5x + 3} \not\equiv -x$.
Also addressing your comment:

The idea behind $\lim_{x \to -\infty} \sqrt{x^2 + 5x + 3} = \dots = \lim_{x \to -\infty} −x$ was, that $x^2$ dominantes $5x+3$. I also thought if that argument is wrong here, if $\lim_{x \to -\infty} x − x = 0$ is legal and everything of the same "type"; then I'm okay again. Then I have to think about why the domination argument doesn't work as I thought it does here.

You are right in thinking that $x^2$ dominates $5x+3$, so that $\color{blue}{\sqrt{x^2 + 5x + 3}} \sim \color{red}{-x}$ for large negative $x$. Indeed there is no problem when computing their ratio:
$$\lim_{x \to -\infty} \frac{\color{blue}{\sqrt{x^2 + 5x + 3}}}{\color{red}{-x}} = 1$$
However, because you are computing the difference between $\color{blue}{\sqrt{x^2 + 5x + 3}}$ and $\color{red}{-x}$, such an approximation is not fine enough. You need more terms of the series expansion:
$$
\begin{align*}
\color{blue}{\sqrt{x^2 + 5x + 3}}
&= -x \sqrt{1 + \frac{5}{x} + \frac{3}{x^2}} \\
&\sim -x \left(1 + \frac{1}{2} \cdot \frac{5}{x} + O\left(\frac{1}{x^2}\right) \right) \\
&= \color{red}{-x} - \frac{5}{2} + O\left(\frac{1}{x}\right)
\end{align*}
$$
Then may you conclude that the required limit is $-5/2$.
By analogy, an approximation such as $\color{blue}{x + 1} \sim \color{red}{x}$ is perfectly fine for computing the ratio limit $\lim_{x \to -\infty} (\color{blue}{x + 1}) / (\color{red}{x})$. But if you are computing the difference limit, then you may not discard the $1$ in $\color{blue}{x + 1}$.
