Proof that $N=49$ is impossible:
+-------+--- 1) +-------+ 2) +-------+ 3) +-------+
| 1 2 3 | X |(1)2 3 |*1 | 1 2 3 |(7) | 1 2 3 |(4)
| 4 5 6 | Y |(4)5 6 |*4 | 4 5 6 | 1 |(4)5 6 | 7
| 7 8 9 | Z |(7)8 9 |*7 |(7)8 9 | 4 | 7 8 9 | 1
+-------+--- +-------+ +-------+ +-------+
| A B C | D | A B C | D | A*B*C |*D | A*B*C |*D
The block on the first figure is to be taken as our "starting" one. Since there are no other numbers on the board, having them ordered makes no difference, for now. Now, there are 3 different ways to fill tiles XYZ with 1 4 and 7, but they all fail. Since:
Case 1: Obviously, there's 3 collisions.
Case 2: BCD must contain 237 on any order. However, 7 can't be on B or C because it collides with our original grid. And it can't be on D either, because it's on the same row as the recently added one.
Case 3: Similar to before, except with 4.
That means there's no way to have those 2 blocks (23X,56Y,89Z and 56Y,89Z,BCD) successfully.
Therefore, $N=49$ is impossible.
Weak continuous Sudoku:
A weak continuous Sudoku can be constructed based on the ideas that you already provided.
First, we construct a weak continuous Sudoku for the set $U=(0,1]$ instead of $U=[0,1]$.
Here, a weak continuous Sudoku can be constructed by using the function
$f$ from your attempt but as a function $f:(0,1]^2\to (0,1]$ (since one boundary is gone, the problems that you observed are now gone, too).
Then, choose a bijection $h:[0,1]\to (0,1]$
(an explicit bijection can be constructed if you prefer a constructive soution).
Then we define
$$
g:[0,1]^2\to [0,1],
\qquad
(x,y)\mapsto h^{-1} (f(h(x),h(y))).
$$
This function $g$ then can be shown to be a weak continuous Sudoku.
Strong continuous Sudoku:
As for strong continuous Sudoku, things get more complicated
and it would be a lot of work to explain my construction in full detail,
but I can provide a sketch.
First, the bijection $h$ above should be chosen such that
each interval in $[0,1]$ contains a subinterval $[ a,b ]$ such that $h(x)=x$
for all $x\in[a,b]$, see the comments below for such a construction.
Furthermore, it uses a bijection $j:[0,1]\to [0,1]$
such that $j((c,d))$ is dense in $[0,1]$ for all intervals $(c,d)$,
see the comments below for such a construction for $j$.
Then one can mix the rows or columns of the previous weak Sudoku according to $j$,
i.e. $\tilde g(x,y)=g(j(x),y)$.
This function $\tilde g$ should then be a strong continuous Sudoku.
Let me provide a rough sketch how this can be done.
Let $S$ be a square sub-region of $[0,1]^2$.
Let $S_2=[a,b]\times [c,d]\subset S$ be a smaller square sub-region,
where $a<b,c<d$ are such that
$h(x)=x$ holds for all $x\in[a,b]\cup[c,d]$
(such a sub-region exists due to the comments above on the choice of $h$).
It suffices to show that $\tilde g(S_2)=[0,1]$ instead of $\tilde g(S)=[0,1]$.
Let $t\in [0,1]$ be given.
Let $m:=(c+d)/2$.
Since $j([a,b])$ is dense in $[0,1]$,
the function values $\{\tilde g(x,m)| x\in[a,b]\}$ are also dense in $[0,1]$.
Let $s\in[a,b]$ be such that $\tilde g(s,m)$ is close to $t$ in the sense that
$$
t-\frac{d-c}{2} < \tilde g(s,m) < t+\frac{d-c}{2}.
$$
By exploiting the definitions of $\tilde g,g,f$ we have
$\tilde g(s,m+x)=\tilde g(s,m)+x$
for $x\in (-\frac{d-c}{2},\frac{d-c}{2})$
(with the exception that the values wrap around at $1$).
By setting $x=\tilde g(s,m)-t$,
we get $t=\tilde g(s,m+x)$ and $(s,m+x)\in S_2 = [a,b]\times [c,d]$.
Thus $t$ can be reached and the condition (5.) for strong continuous Sudoku
is satisfied.
on the existence of a function $h$:
We can define $h:[0,1]\to (0,1]$ by setting $h(0)=1/2$, $h(1/2)=1/3$, $h(1/3)=1/4$,
etc., and $h(x)=x$ for all other $x$.
Then for each interval one can find a sufficiently small subinterval
$[a,b]$ such that $h(x)=x$ for all $x\in[a,b]$.
on the existence of a function $j$:
This is more complicated, so let me provide a rough sketch.
Let $(q_k)_k$ be an enumeration of the rational numbers in $[0,1]$
and let $I_k$ be an interval of length $2^{3-2k}$ centered at $q_k$.
We define the sets
$$ A_k := I_k\setminus \bigcup_{l>k} I_l.$$
These sets form a partition of $[0,1]$ and each set $A_k$ has cardinality
equal to $[0,1]$.
Let $(B_k)_k$ be another sequence of subsets of $[0,1]$ which
form a partition of $[0,1]$ such that each $B_k$ is dense and has
cardinality equal to $[0,1]$
(such a partition exists, one can append dense countable sets with enough other elements to form sets $B_k$,
but I think this requires the axiom of choice).
Then we construct $j$ by (bijectively) mapping
$A_k$ to $B_k$.
Since the lengths of the sets $A_k$ get smaller and smaller
and the rationals $q_k$ are dense,
each interval has a subinterval of the form $I_k$.
Since $I_k$ contains $A_k$ and $A_k$ is mapped to a dense set $B_k$,
we obtain the desired property that $j(I_k)$ is dense in $[0,1]$.
Best Answer
Not all of your $384$ possibilities actually work. For example one of your $384$ choices seems to be
$\begin{matrix} 1&2&|&3&? \\ 3&?&|&2&? \\ -&-&+&-&- \\ 2&3&|& ? & ? \\ ?&?&|& ? & ? \end{matrix}$
forcing
$\begin{matrix} 1&2&|&3&4 \\ 3&4&|&2&1 \\ -&-&+&-&- \\ 2&3&|& 1/4 & ? \\ 4&1&|& ? & 2/3 \end{matrix}$
but it is impossible to complete the grid.
Your argument about $8$ symmetries might fail if there is a pattern which is rotationally symmetric with a $180^\circ$ turn, which there is:
$\begin{matrix} 1&2&|&3&4 \\ 3&4&|&1&2 \\ -&-&+&-&- \\ 2&1&|& 4 & 3 \\ 4&3&|& 2 & 1 \end{matrix}$