The output system is:
$x(t)$ -->(S)--> $y(t) = \int_{-\infty}^{t}x(\tau) d\tau$
Recall that the system is causal if the output at $t$ depends only on input before $t$, or if the impulse response $h(t)=0,\forall t <0$
I guess the system is not causal, but I am not sure how to obtain the impulse response if we are given the output.
On
Note that the input/output relation of your system can be written in terms of a convolution:
$$y(t)=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau\tag{1}$$
where $u(t)$ is the unit step function, which is zero for $t<0$ and equals $1$ for $t>0$. Eq. (1) proves that the system is linear and time-invariant. Furthermore, it proves that the system is causal because the system's impulse response equals the unit step function, which is zero for $t<0$.
Let $ (Lx)(t) = \int_{-\infty}^t x(\tau)d \tau$.
You want to show that if $x_1,x_2$ are such that $x_1(s) = x_2(s)$ for $s \le t$, then $(Lx_1)(t) = (L x_2)(t)$.
You have $(Lx_1)(t) = \int_{-\infty}^t x_1(\tau)d \tau = \int_{-\infty}^t x_2(\tau)d \tau = (L x_2)(t)$.
Hence the system $L$ is causal.
Alternatively, albeit the use of distributions is unnecessary, we can look at the response of the system when subjected to an impulse input. You need to verify that the system is linear and time invariant first, then note that if $t < 0$, we have $h(t) = (L \delta)(t) = 0$.