In the discrete-time case you can use the $\mathcal{Z}$-transform:
$$\mathcal{Z}\{x(n)\}=X(z)=\sum_{n=0}^{\infty}x(n)z^{-n}\tag{1}$$
Equation (1) is the unilateral $\mathcal{Z}$-transform which is useful for causal signals which are zero for $n<0$. Now consider the signal $x(n)=a^n$, $n\ge 0$:
$$X(z)=\sum_{n=0}^{\infty}a^nz^{-n}$$
This series converges for $|a|<1$ and in this case $X(z)$ is given by:
$$X(z)=\frac{1}{1-az^{-1}}=\frac{z}{z-a}\tag{2}$$
From (2) you can see that $a$ is the pole of $X(z)$, and for $x(n)$ to be stable (if seen as a system's impulse response), and for $X(z)$ to exist, $|a|<1$ must hold, i.e. the pole must be inside the unit circle of the complex $z$-plane. This example is analogous to the continuous-time example in your question.
Also take a look at the concept of BIBO-stability (bounded-input-bounded-output stability), which leads to the same conclusion: for continuous time signals/systems the poles must be in the strict left half of the $s$-plane, and for discrete-time signals/systems the poles must be strictly inside the unit circle of the $z$-plane. The condition on the time-domain signals is that they be absolutely integrable in the continuous case, and absolutely summable in the discrete-time case:
$$\int_{-\infty}^{\infty}|x(t)|dt <\infty\\
\sum_{n=-\infty}^{\infty}|x(n)| <\infty$$
