Although my question above concerns whether or not the posted identity is a known result, I'll add the following elementary and direct proof (without recourse to Legendre's Theorem or the de Bruijn function needed for a more general version).
Proposition For real $x \geqslant 1$, the following identity holds \begin{align} \lfloor x \rfloor! = x^{\lfloor x \rfloor} e^{1-x} e^{\int_1^{x} \frac{\text{frac}(t)}{t} \ dt}. \end{align}
Proof. For each positive integer $n$, the function $f(t) = \tfrac{\text{frac}(t)}{t}$ equals $1 - \tfrac{n}{t}$ on $[n,n+1]$. Thus, \begin{align} \int_{1}^{n} f(t) \ d t & = \sum_{k = 1}^{n-1} \int_{k}^{k+1} (1 - \tfrac{k}{t}) \ dt \\ & = n - 1 + \log n! - n \log n. \end{align} By splitting the following integral into two pieces, we compute \begin{align} \int_{1}^{x} f(t) \ d t & =\int_{1}^{\lfloor x \rfloor} f(t) \ d t + \int_{\lfloor x \rfloor }^{x} f(t) \ dt \\ & = \lfloor x \rfloor - 1 + \log \lfloor x \rfloor! - \lfloor x \rfloor \log \lfloor x \rfloor + \int_{\lfloor x \rfloor}^{x } f(t) \ d t. \end{align} We need only to compute the last integral, \begin{align} \int_{\lfloor x \rfloor}^{x } f(t) \ dt & = \int_{0}^{\text{frac}(x) } \frac{\text{frac}(t + \lfloor x \rfloor)}{t + \lfloor x \rfloor} \ d t \\ & = \int_{0}^{ \text{frac}(x) } \frac{t}{t + \lfloor x \rfloor } \ d t \\ & =\text{frac}(x) + \lfloor x \rfloor \log \lfloor x \rfloor - \lfloor x \rfloor \log x \end{align} by a change of variables from $t$ to $t + \lfloor x \rfloor$. Hence, together these identities yield \begin{align} \int_{1}^{x} f(t) \ dt & = x - 1 + \log \lfloor x \rfloor! - \lfloor x \rfloor \log x, \end{align} which completes the proof of the claim.