This is not a complete answer (too big for a comment) but this will help you get started. Define $h_0 = 1$, $h_n : [-1,1]\setminus\{0\} \to \mathbb{N}=\{1,2,\cdots\}$ by $h_n(x) = \text{Abs}(\lfloor x^{-n} \rfloor)$ and take $\epsilon = -\frac{1}{n}$, for $n \in \mathbb{N}$. The definition of $h_n$ can be simplified as following: $$h_n(x) = \begin{cases}k , \,\, n \text{ even and } x \in [-k^{\epsilon},-(k+1)^{\epsilon}) \cup ((k+1)^{\epsilon},k^{\epsilon}], \, k \in \mathbb{N} \\ k , \,\, n \text{ odd and } x \in (-(k-1)^{\epsilon},-k^{\epsilon}] \cup ((k+1)^{\epsilon},k^{\epsilon}], \, k \ge 2 \\
1, \,\, n \text{ odd and } x \in \{-1\} \cup (2^{\epsilon},1] \end{cases}.$$
In terms of $h_n$, the terms of the product can be written as $a_n = \int_{-1}^{1} h_{n-1}(t)/h_n(t) dt $. With the above simplified definition of $h_n$, we can already compute $a_1$, which is, $$a_1 = \int_{-1}^{1} \frac{dt}{h_1(t)} = \int_{1/2}^{1} dt + \sum_{k=2}^{\infty} \frac{1}{k}\left(\int_{\frac{1}{k+1}}^{\frac{1}{k}}+\int_{\frac{-1}{k-1}}^{\frac{-1}{k}}dt\right) = \frac{1}{2}+\sum_{k=2}^{\infty}\frac{2}{k(k^2-1)}=1.$$
However, other terms can't be expressed in closed form, this can be shown by computing $a_2$, $$\begin{gather*}a_2 = \int_{-1}^{1} \frac{h_1(t)}{h_2(t)}dt = \sum_{k=1}^{\infty} \frac{1}{k}\left(\int_{-\frac{1}{\sqrt{k}}}^{-\frac{1}{\sqrt{k+1}}}h_1(t) dt+\int_{\frac{1}{\sqrt{k+1}}}^{\frac{1}{\sqrt{k}}} h_1(t) dt\right) \\
= \sum_{l=1}^{\infty} \sum_{k=l^2}^{(l+1)^2-1} \frac{1}{k}\left((l+1)\int_{-\frac{1}{\sqrt{k}}}^{-\frac{1}{\sqrt{k+1}}} dt + l \int_{\frac{1}{\sqrt{k+1}}}^{\frac{1}{\sqrt{k}}} dt\right) \\
= \sum_{l=1}^{\infty} \sum_{k=l^2}^{(l+1)^2-1} \frac{2l+1}{k}\left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} \right) \approx 1.42722.\end{gather*}$$
To get an estimate of $a_n$, observe that for $n$ even $\frac{1}{x^n}-1 \le h_n(x) \le \frac{1}{x^n}$ and for $n$ odd, $\frac{1}{x^n}-1 \le h_n(x) \le \frac{1}{x^n}, x \in (0,1]$, $-\frac{1}{x^n} \le h_n(x) \le 1-\frac{1}{x^n}, x \in [-1,0)$. For $n$ odd, we have
$$ \int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt- \int_{-1}^{1} \frac{1}{h_n(t)} dt \le a_n \le \int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt$$
where, $$\begin{align*} \int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt &= \int_{2^{\epsilon}}^{1} \frac{1}{t^{n-1}}dt + \sum_{k=2}^{\infty} \frac{1}{k}\left( \int_{-(k-1)^{\epsilon}}^{-k^{\epsilon}} + \int_{(k+1)^{\epsilon}}^{k^{\epsilon}} \frac{1}{t^{n-1}}dt\right) \\
&= \frac{1}{n-2}\left[(2^{1+2\epsilon}-1)+\sum_{k=2}^{\infty} \frac{1}{k}\left( (k+1)^{1+2\epsilon}-(k-1)^{1+2\epsilon}\right) \right]\end{align*}$$
and, $$\int_{-1}^{1} \frac{dt}{h_n(t)} = \int_{2^{\epsilon}}^{1}1 + \sum_{k=2}^{\infty} \frac{1}{k}\left( \int_{-(k-1)^{\epsilon}}^{-k^{\epsilon}} + \int_{(k+1)^{\epsilon}}^{k^{\epsilon}} dt\right) = (1-2^{\epsilon})+\sum_{k=2}^{\infty} \frac{1}{k}\left( (k-1)^{\epsilon}-(k+1)^{\epsilon}\right). $$
The sum in the above expression converges because $(k+1)^{1+2\epsilon}-(k-1)^{1+2\epsilon} \sim (2+4\epsilon)k^{2\epsilon}$ as $k \to \infty$.
Edit 1: Just realized that $\lim_{n\to\infty}\int_{-1}^{1} \frac{1}{h_n(t)} dt = 0$. For $n$ odd case, $$ \begin{align*}0 \le \int_{-1}^{1} \frac{1}{h_n(t)} dt &\le -\int_{-1}^{0} t^n dt + \int_0^{2^{\epsilon}} \frac{t^n}{1-t^n}dt + \int_{2^{\epsilon}}^{1} 1 dt \\
&= \frac{1}{n+1}+\frac{1}{n}\int_{0}^{\frac{1}{2}}\frac{t^{1/n}}{1-t}dt+(1-2^{-\frac{1}{n}}) \end{align*} $$
therefore, ($n$ odd), $$ a_n -\int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt = O\left(\frac{1}{n}\right).$$
Edit 2: If (big if) we can directly substitute the difference $(k+1)^{1+2\epsilon}-(k-1)^{1+2\epsilon} \sim (2+4\epsilon)k^{2\epsilon}$ in the expression for $\int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt$, then the summation is $\zeta\left(1+\frac{2}{n}\right)$ (Reimann zeta function). Taylor expanding this, we obtain that $$ a_n = 1 + O\left(\frac{1}{n}\right).$$ This will imply that the product diverges from this result.
Best Answer
$\newcommand{\d}{\,\mathrm{d}}$A partial answer. We should find: $$\begin{align}I&=\int_0^1\left\{\frac{1}{x\cdot\{\frac{1}{x}\}}\right\}\d x\\&=\sum_{n=1}^\infty\int_{\frac{1}{n+1}}^{\frac{1}{n}}\left\{\frac{1}{x(\frac{1}{x}-n)}\right\}\d x\\&=\sum_{n=1}^\infty\sum_{m=n+1}^\infty\int_{\frac{m-1}{mn}}^{\frac{m}{(m+1)n}}\left(\frac{1}{1-nx}-m\right)\d x\\&=\sum_{n=1}^\infty\frac{1}{n}\sum_{m=n+1}^\infty\left[\ln\left(1+\frac{1}{m}\right)-\frac{1}{m+1}\right]\\&=\sum_{n=1}^\infty\frac{1}{n}\lim_{N\to\infty}[\ln(N)-\ln(n+1)-H_{N+1}+H_{n+1}]\\&=\sum_{n=1}^\infty\frac{H_{n+1}-\ln(n+1)-\gamma}{n}\\&=1-\sum_{n=1}^\infty\frac{\gamma-(H_n-\ln(n+1))}{n}\end{align}$$But I don't believe I know any nice representations for the error $H_n-\ln(n+1)-\gamma$ and don't know how to proceed. I am certain such representations exist, though. I mean, I can cook up some basic integrals but they don't have a nice form. It is nice to observe the numerators in the final form of that series are precisely the differences between the differences in area between the graphs of $1/\lfloor x\rfloor$ and $1/x$ over $[1,n+1]$ and the limit of these differences, $\gamma$. So crudely you could rewrite the summands as $\frac{\gamma}{n}-\frac{1}{n}\int_1^{n+1}\left(\frac{1}{\lfloor x\rfloor}-\frac{1}{x}\right)\d x$ but I doubt that that is useful.