should this be viewed as an impulse response or step response
Rather than spending an efford on the link-exploration, I would instead state here the simplest and the most basic description of the impulse response and step response of causal LTI systems within the particularity of a continuous time leaky integrator. I assume that you know about linear time invariant (LTI) systems and basic differential equations.

A (continuous-time) LTI system is characterised by its impulse response $h(t)$ defined via the relationship $$h(t) = \mathcal{T} \{ \delta(t) \}$$ such that
$$ \delta(t) \rightarrow \boxed{ h(t) } \rightarrow h(t)$$

Where $\delta(t)$ is the impulse that is at the input. The step response can be defined as $$s(t) = \mathcal{T} \{ u(t) \}$$ such that
$$ u(t) \rightarrow \boxed{ h(t) } \rightarrow s(t)$$ where $u(t)$ is the unit-step function.

One can further relate the step response to impulse response as $$ s(t) = \int_{-\infty}^{t} h(\tau) d\tau \longleftrightarrow h(t)=s'(t)$$

Finally, for an arbitrary input $x(t)$ the corresponding output $y(t)$ of any LTI system is given by the convolution integral as: 
$$ y(t) = h(t) \star x(t) = \int_{-\infty}^{\infty} h(t-\tau) x(\tau) d\tau = \int_{-\infty}^{\infty} x(t-\tau) h(\tau) d\tau = x(t) \star h(t) $$

Now before coming to the leaky-integrator, we shall consider the ideal integrator:

The ideal integrar is the system with the input-output relationship as
$$ y(t) = \mathcal{T}\{ x(t) \} = \int_{-\infty}^{t} x(\tau) d\tau $$

Considering that ideal-integrator is an LTI system we can regard the above integral as a convolution operator and hence deduce that the impulse response of the ideal integrator is $$h(t) = u(t)$$ then the step response will be $$s(t) = \int_{-\infty}^t u(\tau) d\tau = t u(t) = r(t)$$ $r(t)$ being the ramp function.

Note that the differential equation associated with the ideal-integrator can be obtained by diferentiating its input-output relationship with respect to time as

$$ \frac{d y(t) }{dt} = \frac{d}{dt} \int_{-\infty}^t x(\tau)d\tau $$ which results in (after following the differentiation of the integral) 
$$ y'(t) = x(t) $$ with properly chosen initial conditions (according to the applied input) so that the ODE (an LCCDE) represents an LTI system. 

The leaky-integrator can be defined in a number of ways as the term leaky may accept multiple mathematical definitions but one of the most common has the following input-output relationship:

$$y(t) = \mathcal{T} \{x(t) \} = \int_{-\infty}^{t} e^{-\alpha (t-\tau)} x(\tau) d\tau $$ Which has the interpretation that, when $\alpha &gt; 0$, past values of the input are ignored more and more as time goes on, when computing the current value of the output. This is achived by the decaying-exponential weighting applied to the integrator input $x(t)$.

Again considering that this is an LTI system and therefore that its output is given by a convolution integral, then we can deduce that the impulse response $h_l(t)$ of the leaky integrator is 
$$h_l(t) = e^{-\alpha t} u(t)$$ and the associated step response can be obtained as $$s_l(t) = \int_{-\infty}^t h_l(\tau)d\tau = \int_{-\infty}^t e^{-\alpha \tau} u(\tau) d\tau = \int_{0}^t e^{-\alpha \tau} d\tau = \frac{1}{\alpha} (1-e^{-\alpha t})$$   

The associated ordinary differential equation of the leaky-integrator can be obtained similarly by diferentiating its input-output relationship with respect to time $t$ as

$$ \frac{d y(t) }{dt} = \frac{d}{dt} \int_{-\infty}^{t} e^{-\alpha (t-\tau)} x(\tau) d\tau .$$ After following the differentiation of the integral carefully one can reach
$$ y'(t) = e^{-\alpha (t-t)}x(t) t' - e^{-\alpha(t+\infty)}x(-\infty) \frac{d}{dt}(-\infty) + \int_{-\infty}^{t} -\alpha e^{-\alpha (t-\tau)} x(\tau) d\tau  $$ 
It can be seen that the first term is $x(t)$, the second term is zero and the last term is $-\alpha y(t)$, hence  we deduce
$$y'(t) = x(t) - \alpha y(t) \longleftrightarrow y'(t)+\alpha y(t) = x(t)$$ as the ODE that define the LTI leaky integrator.