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2 added 130 characters in body

Some clarifications: 1) You need that $X$ is grouplike (so the induced multiplication makes $\pi_0 X$ a group). This condition is always satisfied for a loop space, but not satisfied by the discrete $E_\infty$ space $\mathbb{N}$.

2) In order for $\mathcal{L}$ to be an $E_\infty$ operad we require more than $\mathcal{L}(n)\simeq *$, otherwise we could have just used the commutative operad and all grouplike $E_\infty$ spaces would be topological abelian groups. We also require that $\mathcal{L}(n)$ is a cofibrant $\Sigma_n$-space, so that it has a free $\Sigma_n$ action and consequently homotopy equivalent to $E\Sigma_n$.

In May's 'Geometry of iterated loop spaces' especially Ch. 3 http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf it is shown that the delooping machine does not depend on the choice of $E_\infty$ operad.

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Some clarifications: 1) You need that $X$ is grouplike (so the induced multiplication makes $\pi_0 X$ a group). This condition is always satisfied for a loop space, but not satisfied by the discrete $E_\infty$ space $\mathbb{N}$.

2) In order for $\mathcal{L}$ to be an $E_\infty$ operad we require more than $\mathcal{L}(n)\simeq *$, we also require that $\mathcal{L}(n)$ is a cofibrant $\Sigma_n$-space, so that it has a free $\Sigma_n$ action and consequently homotopy equivalent to $E\Sigma_n$.

In May's 'Geometry of iterated loop spaces' especially Ch. 3 http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf it is shown that the delooping machine does not depend on the choice of $E_\infty$ operad.