Let $f(x), g(x) \geq 0$. Then, I want to know if the following is true
$$ \arg \max_x [f(x)g(x)] = \arg\max_x f(x) \cdot \arg\max_x g(x) $$
And how one can prove it.
I found a related question in this post but neither the notation nor the answer were clear enough for me.
There are some serious differences between your question and the linked question. The linked question is indeed accurate, once you unpack the notation a little.
Explaining The Linked Question
The linked question is about maximising $3$ different function $f$, $g$, $h$, each with their own independent variable, over the respective domains of the functions. There is no need to assume that the domains are subsets of $\Bbb{R}$; they can be any three sets $X, Y, Z$, that may have nothing to do with each other.
The left hand side $$\DeclareMathOperator*{\argmax}{argmax} \argmax_{x,y,z} f(x) \cdot g(y) \cdot h(z)$$ refers to the set of points $(x, y, z) \in X \times Y \times Z$ such that $f(x)g(y)h(z)$ attains its maximum value. The right hand side is a product of argmaxes such as $\argmax_x f(x)$, which, in this case, is the set of points in $X$ that maximise $f(x)$. From context, this means the product between the sets on the right hand side is a Cartesian product, not a product of real numbers.
As a concrete example, let's define two functions (instead of three): $f : [-1, 1] \to [0, \infty)$ and $g : \Bbb{R} \to [0, \infty)$, where $f(x) = 1 - x^2$ and $g(y) = 1 + \cos(y)$. Then, I know that \begin{align*} \argmax_{x \in [-1, 1]} f(x) &= \{0\} \\ \argmax_{y \in \Bbb{R}} g(y) &= 2\pi\Bbb{Z} = \{\ldots, -4\pi, -2\pi, 0, 2\pi, 4\pi, \ldots\} \\ \argmax_{(x, y) \in [-1, 1] \times \Bbb{R}} f(x)g(y) &= \left(\argmax_{x \in [-1, 1]} f(x)\right) \times \left(\argmax_{y \in \Bbb{R}} g(y)\right) \\ &= \{0\} \times 2\pi\Bbb{Z} \\ &= \{\ldots, (0,-4\pi), (0,-2\pi), (0,0), (0,2\pi), (0,4\pi), \ldots\}. \end{align*}
That is, the function $(x, y) \mapsto f(x)g(y)$ achieves its maximum on the above set (which, in this case, is a subset of the Cartesian plane).
Your Question
While your notation is not 100% clear here, I imagine that $f(x)$ and $g(x)$ are real functions. This makes $f(x)g(x)$ a real function, and the argmax a subset of $\Bbb{R}$. This means that the right hand side cannot be a Cartesian product, as in the linked question. Instead, I imagine you're interpreting it using the Minkowski product: $$A \cdot B = \{ab : a \in A, b \in B\},$$ i.e. the set of all products you can form by taking one operand from each set. If $A = \{a\}$ for some $a$ and $B = \{b\}$ for some $b$, i.e. if the sets contain only one element, then naturally, the Minkowski product is $\{ab\}$. That is, if $f(x)$ and $g(x)$ are maximised uniquely at $x = a$ and $x = b$ respectively, then I imagine that you expect $f(x)g(x)$ to be uniquely maximized at $x = ab$.
This is not true. While plugging $x = a$ or $x = b$ into $f(x)$ or $g(x)$ respectively produces something of note, plugging in this third value $x = ab$ will produce nothing special.
As a concrete example, take $f(x) = 1 + \cos x$ and $g(x) = 1 - \cos x$, and to make things easier, consider it over the domain $[0, \pi]$. Then $$\argmax_x f(x) = \{0\}, \quad \argmax_y g(x) = \{\pi\}.$$ But, their product $f(x)g(x) = 1 - \cos^2(x) = \sin^2(x)$ is maxmised at $\pi/2$, i.e. $$\argmax_x f(x)g(x) = \left\{\frac{\pi}{2}\right\},$$ whereas $$\left(\argmax_x f(x)\right) \cdot \left(\argmax_x g(x)\right) = \{0 \cdot \pi\} = \{0\},$$ making the statement untrue in general.
Here's the best I can do in this direction. I know that if both functions happen to be maximised at the same time, then the product will be maximised! That is,
$$\left(\argmax_x f(x)\right) \cap \left(\argmax_x g(x)\right) \subseteq \argmax_x f(x) g(x).$$