Suppose that
$$\mathrm A = \mathrm O_2 \qquad \qquad \mathrm b = \begin{bmatrix} 0\\ 1\end{bmatrix} \qquad \qquad \mathrm c = \begin{bmatrix} 1 & 1\end{bmatrix}$$
and that the initial condition is $\mathrm x_0 := (x_{10}, x_{20})$. Hence, the states are
$$x_1 (t) = x_{10} \qquad \qquad \qquad x_2 (t) = x_{20} + \int_0^{t} u (\tau) \,\mathrm{d}\tau$$
and the output signal is
$$y (t) = (x_{10} + x_{20}) + \int_0^{t} u (\tau) \,\mathrm{d}\tau$$
Suppose that we start from zero initial conditions and that the input signal is constant, say, $u = 1$. Hence, the output signal is given by
$$y (t) = \int_0^{t} u (\tau) \,\mathrm{d}\tau = t$$
Thus, even though the LTI system is internally marginally stable (the zero eigenvalue has two $1 \times 1$ Jordan blocks, rather than one $2 \times 2$ Jordan block), it is not BIBO (bounded input, bounded output) stable, as a bounded input can produce an unbounded output.
First, consider the following first order transfer function:
$$ \frac{X(s)}{U(s)} = \frac{a}{s - a} $$
where $a \in \mathbb{C}$ is the system pole. If we observe the behavior of the system in time we have
$$ \dot{x}(t) = e^{a t} (u(t) - x(t))$$
Since $ a $ is complex we can write it as $ a = b + j c $ where $ b $ is the real part of $ a $ and $ c $ the imaginary part. Then the system becomes:
$$ \dot{x}(t) = e^{b t} e^{j c t} (u(t) - x(t))$$
Note that $ e^{j c t} $ will cause the system to oscillate, while $ e^{b t} $ will determine how (and if) the $x$ will converge to $u$.
If $ b < 0 $ the system will go to zero since $ e^{b t} \rightarrow 0 $ when $ t \rightarrow \infty $. Meanwhile, if $ b > 0 $ the system will diverge since $ e^{b t} \rightarrow \infty $ when $ t \rightarrow \infty $.
Note that $ c $ does not play a role here. So independently of the imaginary part, the real part of the pole needs to be negative for stability.
For the case $ b = 0 $ the system will neither converge nor diverge, however stability is defined by strict convergence, so $ b = 0 $ is not stable (attention to the choice of words, it may or may not be unstable, I recommend you to read more on marginal stability for this).
Best Answer
The definition as stated on the Wikipedia page of Minimum phase reads as follows:
A causal system means that the system doesn't rely on future inputs for the current state of the system (more zeros than poles to put it really easily for now). And (BIBO) stable meaning that the states are bounded or $\forall t:|x(t)|<\infty$. Note that this doesn't require that the states to converge to zero (which should be the case if you're requiring a-symptotic stability).
Having a system with for example the transfer function $H(s)=\frac{s^2+10}{s^2+1}$ satisfies both these criteria, as well as that the inverse satisfies these criteria, meaning a minimum phase system, but it's easy to see that in both cases the poles of $H(s)$ and $H_{inv}(s)$ have poles on the imaginary axis.
(in case of $H(s)$, $p_{1,2}=+/- i$, and in case of $H_{inv}(s)$, $p_{1,2}=+/- i\sqrt{10}$.)
The physical meaning of this is that the system oscilates around a specific point with a certain frequency.
Always try to think of solutions of systems in exponential powers (since systems are written as differential equations, and the solutions to differential equations normally are in the form of exponential powers, in state-space description $\dot{x}(t)=Ax$ has as a solution $x(t) = e^{At}$). This is also why you want your poles to be in the left half plane, such that you can have solutions of the form $x(t) = e^{p t}$, where if $p<0$ the exponent converges to zero, and if $p$ is complex/purely imaginary you can use Euler's formula to rewrite the solution as sines and cosines.
Which is the case if you have purely imaginary poles you get solutions of the form $e^{ict} = c (\cos(t)+i \sin(t)$.