How to prove that : the graph of function $f$ is symmetric w.r.t. point $I=(a,b)$ iff $ f(2a-x)+f(x)= 2b$. Source : https://www.youtube.com/watch?v=3RHiaIKJeBA , claiming that function $f$ is symmetric w.r.t. a point $I=(a,b)$ iff $ f(2a-x)+f(x)= 2b$.

Symmetry with respect to $O=(0,0)$ is defined by $ f(-x)= -f(x)$.
Saying that function $f$ is symmetric with respect to a point $I=(a,b)$ is equivalent to saying that in a cartesian coordinate system centered at $I$, function $g(x)= f(x-a) + b$ is symmetric with respect to the origin.
So we should have  $g(-x)= -g(x)$, that is $ f(-x-a)+b = -(f(x-a)+b) =  -f(x-a) -b$.
Manipulating this last expression, I get :
$ f(-x-a)+b = -f(x-a)-b$
$\iff f(-x-a) +f(x-a) = -2b$
$\iff -f(-x-a) -f(x-a) = 2b$
Not quite identical to the desired formula.

You have the right idea. However, note that to get the new function $g(x)$ to be symmetric with respect to the origin means we need to shift the coordinates left by $a$ and down by $b$, so it'll actually be
$$g(x) = f(x + a) - b \tag{1}\label{eq1A}$$
i.e., you have the opposite signs for the $a$ and $b$ terms (with \eqref{eq1A} giving $g(0) = f(a)-b = b - b = 0$, as required). Using the general $g(-y) = -g(y)$, similar to what you did, we next have
$$\begin{equation}\begin{aligned}
f(-y+a)-b & = -(f(y+a) - b) \\
f(-y+a) - b & = -f(y+a) + b \\
f(-y+a) + f(y+a) & = 2b
\end{aligned}\end{equation}$$
Using $x = y+a \; \to \; -y = a -x \; \to \; -y+a = 2a -x$, we then get
$$f(2a-x) + f(x) = 2b$$
as the linked video claims. Note this gives the $\implies$ part of the claim, with the $\impliedby$ part coming from that the actions on the expressions above are all reversible.
