I tried to make sense of the product integrals as given by the OP and the link he provided. I can prove the statement for functions $f$ that are Riemann integrable in some box $C=[a,b]\times[c,d]\subset\mathbb{R}^2$ as follows.
Let $\mathcal{P}:=\{[t_i,t_{i+1}]\times[s_j,s_{j+1}]: 0\leq I< n, 0\leq j< m\}$ be a partition of $[a,b]\times [c,d]$ ($a=t_0<\ldots<t_n=b$, $c=s_0<\ldots<s_m=d\}$. Assume $|f(\mathbf{x})|\leq M$, for all $\mathbf{x}\in C$. Then for any $A\in\mathcal{P}$, choose a tag $\mathbf{x}_A\in A$ and define
$$Q_{\mathcal{P}}=\prod_{A\in\mathcal{P}}\big(1+f(\mathbf{x}_A)\lambda_2(A)\big)$$
where $\lambda_2([t_i,t_{i+1}]\times[s_j,s_{j+1}])=(t_{i+1}-t_i)(s_{j+1}-s_j)$ if $A=[t_i,t_{i+1}]\times[s_j,s_{j+1}]$.
Suppose partitions $\mathcal{P}_k$ are taken so that $\max_{A\in\mathcal{P}_k}\lambda_2(A)=0$.
Then
$\max_{A\in\mathcal{P}_k}|f(\mathbf{x}_A)\lambda_2(A)|\xrightarrow{k\rightarrow\infty}0$
$\lim_k\sum_{A\in\mathcal{P}_k}f(\mathbf{x}_A)\lambda_2(A)=\int_Cf$
$\sup_k\sum_{A\in\mathcal{P}_k}|f(\mathbf{x}_A)|\lambda_2(A)\leq M\lambda_2(C)$
Then, $\lim_kQ_{P_k}=\exp\Big(\int_Cf\Big)$. This follows from Theorem A below. The result can be extended to Riemann integrable functions over bounded closed boxes on $\mathbb{R}^d$. For more general functions, for example, Lebesgue integrable the statement in the OP may not hold.
Theorem A:
Let $\{c_{n,m}:1\leq m\leq m_n\}\subset\mathbb{C}$. Suppose that
- $\lim\limits_{n\rightarrow0}\sup_{1\leq m\leq m_n}|c_{n,m}|=0$,
- $\lim\limits_{n\rightarrow\infty}\sum^{m_n}_{m=1}c_{n,m}=c\in\mathbb{C}$,
- and $M:=\sup_n\sum^{m_n}_{m=1}|c_{n,m}|<\infty$.
Then
\begin{align}\prod^{m_n}_{m=1}(1+c_{n,m})=e^c
\end{align}
A proof pot this result can be found here