Is a leaky integrator the same thing as a low pass filter?
A so-called leaky integrator is a first-order filter with feedback. Let's find its transfer function, assuming that the input is $x(t)$ and the output $y(t)$:

$$
\frac{dy(t)}{dt} + Ay(t) = x(t)
$$

$$
\mathcal{L}\left\{\frac{dy(t)}{dt} + Ay(t)\right\} = \mathcal{L}\left\{x(t)\right\}
$$

where $\mathcal{L}$ denotes application of the Laplace transform. Moving forward:

$$
sY(s) + AY(s) = X(s)
$$

$$
H(s) = \frac{Y(s)}{X(s)} = \frac{1}{s + A}
$$

(taking advantage of the Laplace transform's property that $\frac{dy(t)}{dt} \Leftrightarrow sY(s)$, assuming that $y(0) = 0$).

This system, with transfer function $H(s)$, has a single pole at $s = -A$. Remember that its frequency response at frequency $\omega$ can be found by letting $s=j\omega$:

$$
H(j\omega) = \frac{1}{j\omega + A}
$$

To get a rough view of this response, first let $\omega \to 0$:

$$
\lim_{\omega \to 0} H(\omega) = \frac{1}{A}
$$

So the system's DC gain is inversely proportional to the feedback factor $A$. Next, let $w \to \infty$:

$$
\lim_{\omega \to \infty} H(\omega) = 0
$$

The system's frequency response therefore goes to zero for high frequencies. This follows the rough prototype of a lowpass filter. To answer your other question with respect to its time constant, it's worth checking out the system's time-domain response. Its impulse response can be found by inverse-transforming the transfer function:

$$
H(s) = \frac{1}{s+A} \Leftrightarrow e^{-At}u(t) = h(t)
$$

where $u(t)$ is the Heaviside step function. This is a very common transform that can often be found in tables of Laplace transforms. This impulse response is an exponential decay function, which is usually written in the following format:

$$
h(t) = e^{-\frac{t}{\tau}}u(t)
$$

where $\tau$ is defined to be the function's time constant. So, in your example, the system's time constant is $\tau = \frac{1}{A}$.