How to determine if the system is invertible
It depends on what exactly you mean by "invertible". In system theory, what is often meant is if there is a causal and stable system that can invert a given system, because otherwise there might be an inverse system but you can't implement it.

For linear time-invariant systems there is a straightforward method, as mentioned in the comments by Robert Bristow-Johnson. First, note that the input-output relationship of an LTI system is given by the convolution integral

$$y(t)=\int_{-\infty}^{\infty}h(t-\tau)x(\tau)d\tau\tag{1}$$

Comparing $(1)$ to the system given in your question shows that the system is LTI with impulse response

$$h(t)=e^{-t}u(t)\tag{2}$$

where $u(t)$ is the unit step function. The corresponding transfer function is

$$H(s)=\frac{1}{1+s}\tag{3}$$

This system is causal and stable. However, its inverse system

$$G(s)=\frac{1}{H(s)}=1+s\tag{4}$$

is not stable. It has a pole at infinity. The input-output relation of the inverse system is

$$y(t)=x(t)+x'(t)\tag{5}$$

where $x'(t)$ is the derivative of $x(t)$. So in order to implement the inverse system you need a differentiator, which is not stable in the bounded-input-bounded-output (BIBO) sense. I.e., bounded input signals can lead to unbounded output signals (just use a rectangular input function as an example).