Lets say I have a transfer function $H(s)$ of a system defined in $s$-domain as: $$H(s) = \frac{1}{s - (-1-j)}$$
So I conclude that the pole on the $s$-plane is where $s = 1+j$. So far so good.
Now does that mean if the Laplace transform of the input to the system is $s = 1+j$ the system goes crazy/infinity/oscillate?
Does that mean if I take the inverse Laplace transform of $1+j$ I can find what input in time domain makes the system unstable?
Let $H(s)$ be a transfer function of the form $$H(s) = \frac{1}{s-p}$$ where $p$, which is a pole of $H(s)$, can be written as a complex number $a+jb$. Taking the inverse Laplace transform of $H(s)$ gives the corresponding impulse response $h(t)$ (that is, the output of your system when given $\delta(t)$ as input). Noting $\mathcal{L}^{-1}$ the inverse Laplace transform, we have $$h(t) = \mathcal{L}^{-1}\{H(s)\} = e^{pt} = e^{at}e^{jbt}.$$ Now let's look at what this impulse response looks like. The term $e^{at}$ is a simple exponential which will be either decaying (if $a < 0$) or growing (if $a$ > 0) with time. The term $e^{jbt}$ will be responsible for oscillations in the output of your system (remember that $e^{jbt} = \cos(bt) + j\sin(bt)$). From this, you can infer the stability of your system and understand why we need poles in the left-hand side of the $s$-plane (i.e. we need $a < 0$) for the system to be stable.
Often, the numerator and the denominator of your transfer function have real coefficients, and in this case poles appear in complex conjugate pairs. You could for example have $$h(t) = e^{at}(e^{jbt} + e^{-jbt}) = 2e^{at}\cos{bt}.$$
I like to keep this picture in mind (taken from here) which greatly summarizes this.
For more complex transfer function, partial fraction decomposition can be used to go back to simple cases as presented here.
If
$H(s)= \dfrac{1}{s - (-1-j)}$
then your pole is at $p_1 = -1-j$, which is in the left-hand side of the s plane, so the system is stable.
