Let $D_e$ denote the duplicates of $e$. Let $C_s$ denote the copies of $s$.

Given any fractional set cover $x'$ of ${\cal I}'$, define fractional multicover $x=f(x')$ of ${\cal I}$ by taking $x_s = \sum _{s'\in C_s} x'_{s'}$ to be the sum of the sizes of the copies of $s$ in $x'$. Then $x$ is feasible if $x'$ is, because, for any element $e$, for each of the $d_e$ duplicates $e'$ of $e$, the weight of sets $s'$ containing $e'$ must be at least 1, and each of these sets $s'$ is a copy of $s$:

The cost of $x$ equals the cost of $x'$, and, if $x'$ is an integer solution, then so is $x$.

Conversely, given any fractional multicover $x$ of ${\cal I}$, define fractional set cover $x'$ of ${\cal I}'$ to be an inverse of $f$ (so that $f(x') = x$), computed as follows.

	input: fractional multicover $x$ of ${\cal I}$
	output: fractional set cover $x'$ of ${\cal I}'$
1.	Initialize $x'_{s'} = 0$ for all $s'$.
2.	For each set $s$:
3.	Repeat until $x_s = 0$:
4.	Shift weight from $x_s$ to $x'_{s'}$, where $s'$ is any copy of $s$ such that, for each $e\in s$, if some copy of $e'$ has $x'(e') \lt 1$, then one such $e'$ is in $s'$. (Shift weight to $x'_{s'}$ only as long as the condition on $s'$ continues to hold.)
5.	Return $x'$.

The cost of $x'$ is the same as the cost of $x$, and, if $x$ is an integer solution, then $x'$ will be too (provided the shifts in line 4 are maximal).

If $x$ is feasible, then so is $x'$, because a total of $d_e$ units of weight must be shifted to sets containing copies of $e$, there are only $d_e$ such copies, and no copy $e'$ has $x'(e')$ greater than 1 unless all copies are covered.

Running the localized rounding scheme on a fractional solution $x$ of ${\cal I}$ and returning $\tilde x$ corresponds to running the localized rounding scheme for Set Cover on the fractional solution $x'=f^{-1}(x)$ on ${\cal I}'$, where $f$ is as described in the reduction above, then returning $\tilde x = f(\tilde x')$, where $\tilde x'$ is the rounded solution for ${\cal I}'$. The cost of $\tilde x$ equals the cost of $\tilde x'$. We know that the expected cost of $\tilde x'$ is at most $\sum _{s'\in {\cal I}'} {\rm H}(|s'|) c_s x'_{s'}$ (proven here), which equals $\sum _{s\in {\cal I}} {\rm H}(|s|) c_s x_{s}$.

Running that greedy algorithm on a fractional solution $x$ of ${\cal I}$ and returning $\tilde x$ corresponds to running the greedy algorithm for Set Cover on the fractional solution $x'=f^{-1}(x)$ on ${\cal I}'$, where $f$ is as described in the reduction above, then returning $\tilde x = f(\tilde x')$, where $\tilde x'$ is the rounded solution for ${\cal I}'$. The cost of $\tilde x$ equals the cost of $\tilde x'$. We know that the cost of $\tilde x'$ is at most $\sum _{s'\in {\cal I}'} {\rm H}(|s'|) c_s x'_{s'}$ (proven here), which equals $\sum _{s\in {\cal I}} {\rm H}(|s|) c_s x_{s}$.