Output signal as convolution of impulse response and step signal
It is a normal IIR filter, and a very simple one at that.

Everything that can be described with a linear time invariant difference equation is an IIR filter and can be designed and analyzed using well established mathematical methods.

You calculate the output of the filter for ANY given input (impulse, step, the Dow Jones, Wagner's Nibelungen, etc.) with the equation
$$y[n] =\sum_{i=0}^{N}b_i\cdot x[n-i]-\sum_{i=1}^{N}a_i\cdot y[n-i]$$

In your case you have $b_0=0.2$ and $a_1 = -0.8$ with all the other a and b coefficients being zero, so the equation simplifies to
$$y[n] = 0.2 \cdot x[n] + 0.8 \cdot y[n-1] $$

That is indeed a simple first order low pass filter with a -3dB point at a normalized frequency of 0.0355 or thereabouts. It's not a very good low pass filter since it doesn't go to zero at the Nyquist frequency. A somewhat better choice would be 

$$y[n] = 0.1 \cdot x[n] + 0.1 \cdot x[n-1] + 0.8 \cdot y[n-1] $$

There are plenty of smoothing algorithms that are not IIR filters but your specific example is just a simple first order low pass.

EDIT: based on the question in the comment
The equation above describes the operation for ALL input signals, not just an impulse. 
Let's consider the input signal x = [1.0 1.1 1.05]; So we have x[0] = 1.0, x[1] = 1.1, x[2] = 1.05, and x[3...] = 0; You would calculate the output as follows

y[0] = 0.2*x[0] + 0.8*y[-1] = 0.2; // since y[-1] = 0;
y[1] = 0.2*x[1] + 0.8*y[0]  = 0.38;
y[2] = 0.2*x[2] + 0.8*y[1]  = 0.514;
y[3] =            0.8*y[2]  = 0.4112;     // since x[3] = 0;
y[4] =            0.8*y[3]  = 0.32896; 
etc.