If $\phi\in C_C^\infty$, then we have
$$\begin{align}
\langle H',\phi \rangle &=-\langle H, \phi'\rangle \tag1\\\\
&=-\int_{-\infty}^\infty H(x)\phi'(x)\,dx\tag2 \\\\
&=-\int_{-\infty}^\infty \lfloor x \rfloor \phi'(x)\,dx\tag3 \\\\
&=-\sum_{n=-\infty}^\infty \int_n^{n+1}\lfloor x \rfloor \phi'(x)\,dx\tag4\\\\
&=\sum_{n=-\infty}^\infty n(\phi(n)-\phi(n+1))\tag5\\\\
&=\underbrace{\sum_{n=-\infty}^\infty (n\phi(n)-(n+1)\phi(n+1))}_{\text{Telescopes to }\,0}+\sum_{n=-\infty}^\infty \phi(n+1)\tag6\\\\
&=\sum_{n=-\infty}^\infty \phi(n)\tag7
\end{align}$$
Therefore, in distribution we assert that the floor function is equal to a Dirac Comb given by
$$\bbox[5px,border:2px solid #C0A000]{\lfloor x\rfloor =\sum_{n=-\infty }^\infty \delta(x-n)}$$
NOTES:
The equality in $(1)$ is the definition of the distributional derivative.
Since $H\phi'\in L^1$, the distribution in $(2)$ is an integral over $\mathbb{R}$.
In going from $(2)$ to $(3)$, we used $H(x)=\lfloor x\rfloor$.
In arriving at $(4)$ we divided the integral into the sum of integrals on intervals for which $\lfloor x\rfloor$ is constant.
In going from $(4)$ to $(5)$, we integrated $\phi'(x)$ on the interval $[n,n+1]$.
We simply write $n(\phi(n)-\phi(n+1))=(n\phi(n)-(n+1)\phi(n+1)) +\phi(n+1)$ to obtain $(6)$.
In going from $(6)$ to $(7)$, we summed the telescoping series and used the fact that $\phi\in C_C^\infty$ so that $\lim_{n\to \pm \infty}\phi(n)=0$. We also shifted the index from $n$ to $n-1$.
Best Answer
Using two facts valid in $\mathcal{D}',$
we get that $(S*\rho_n)' = S'*\rho_n \to S'$ in $\mathcal{D}'.$