Why does an argument similiar to 0.999...=1 show 999...=-1?
If you want to understand the mathematics behind these things, it is all based upon the notions of 'convergence' and of 'limits'. If you read any first course in analysis textbook you will find the concept rigorously treated there.
Basically this is the point: Whenever you write 0.999... you are writing down a numeral that represents the 'limit' obtained when an infinite summation $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...$ is performed. Since we can prove that this sum 'converges' to some real number (namely 1), we are justified in treating the numeral 0.999... as representing some real number.
However, whenever you write down 999... I presume you are writing a numeral to represent the limit obtained when an infinite summation $9+90+900+...$ is performed. Since this limit does not converge to any real number, (it 'diverges'), we are not justified in treating the numeral 999... as any real number. So it does not make sense to divide it by ten, or take it away from itself.
We usually denote such divergent limits by the numeral $\infty$, but this does not denote a real number, and there is no consistent way to define operations such as $\infty - \frac{1}{10}\infty$.
I hope this helps and I hope you are motivated to think more about these things :)
In the $10$-adic numbers it is true that $\dots 9999 = -1$.
More precisely, the series $\sum_{n=0}^{\infty} 9 \cdot 10^n$ converges in $\mathbb{Q}_{10}$ and its limit there is $-1$.
As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike the sequence $0.9, 0.99, 0.999, \ldots$, which converges to $1$.
However, you have touched on something interesting. Are there number systems where this does converge? And if so, does it converge to $-1$? The answer to both is yes.
For an integer $k$, there's a ring called the $k$-adic numbers*. Let's use $k = 10$. It's similar to the regular real numbers in base $k$, but a few things are backwards.
$$ \small{\begin{array}{c|c|c} & \textrm{Reals (base }k\textrm{)} & k\textrm{-adic numbers} \\ \hline \textrm{number of digits left of the point} & \textrm{finitely many} & \textrm{infinitely many} \\ \textrm{number of digits right of the point} & \textrm{infinitely many} & \textrm{finitely many} \\ \textrm{two numbers are close if} & \textrm{they match in the leftmost digits} & \textrm{they match in the rightmost digits} \end{array}} $$
That's not really the way that people like to construct the $k$-adics, but it's easier than all that "completion wrt a metric" or "inverse limit" stuff.
Anyways, in this system, $9, 99, 99, \ldots$ does have a limit, and it's $\ldots 999.0$. And if you add $1$ to that, you have the sequence $10, 100, 1000, \ldots$, which converges to $0$. So in that system, $\ldots 999 = -1$.
*Usually people like $k$ to be prime, so they call it $p$. So if you wanna find out more, look up "$p$-adic numbers", not "$k$-adic numbers".