Requirement of time step in convolution output?
The distinction is whether your x-axis is in units of time or units of samples, and if we want our digital system to remain constant with time independent of the sampling rate. Once our system is discrete we often prefer to work in (normalized) units of samples and not have to carry on the extra "step" term, in which case actual time will scale along with the sampling rate (this has been the more common approach in my experience, unless I am working on a mixed signal design). 

A very easy way to see this is to consider a digital integrator (accumulator). We could follow the same process with convolution but using an integrator structure will be much easier, and clearer. First, the graphic below shows how a integrator which has the Laplace Transform $H(s) = \frac{1}{s}$ in the continuous time domain, can be mapped to $H(z) = \frac{T}{1-z^{-1}}$ in the discrete-time domain (using the mapping method of impulse invariance). Notice I stated specifically the discrete-time domain, meaning even though we are working with a discrete system, we still want to work in units of time (seconds). Therefore we must use the $T$ factor in the equation (which is "step" in your question).  If we work in units of samples, then T=1 and essentially disappears, simplifying our math. 



So now the easy part: regardless of all the math and explanation above using "method of impulse invariance" etc let's just believe for a moment that the accumulator structure (keep adding the input to an accumulated value) is a discrete approximation of an integrator. If we add the factor T (or step) then the accumulator is a discrete-time approximation of an integrator. 

Knowing that let's see it in action with reference to the graphic below showing the accumulator input and output.

If our horizontal axis is time (not samples) and if we want a system that keeps the same time performance independent of the sampling rate (meaning we achieve the exact same result with regards to time, the resulting waveforms throughout the system are just sampled faster or slower but the values vs time do not change), then in this case we must include the T factor.  

Consider the simple case of the accumulator where the input is all ones. In discrete-time, regardless of the sampling rate an integration of a constant 1 at the input would grow to 1 after 1 second at the output. If the sampling rate was 10 Hz (for example) then T = 1/10Hz = 0.1. At 10 Hz, there would be 10 samples after 1 second, and without the T factor the output would grow to 10. By using T = 0.1, we get the result of 1 as expected at 1 second.  Change the sampling rate to 100 Hz, and after 1 second the output would grow to 100, but if we add the proper scaling T = 0.01, we still get the same result of 1 as expected at 1 second.   



So in summary we see by including step (or "T" as in my figure) we produce results that are consistent with time (in seconds) regardless of the sampling rate. It is more common to work in units of samples (not seconds) and know that the system will scale along with the sampling rate, but using the units of time has convenience for mixed signal designs - or when a result must be reported in units of time.  We see this with normalized frequency for example where the frequency axis extends from 0 to 1 (cycles/sample) instead of 0 to $f_s$ (cycles/second) as described in this post:

What is normalized frequency