Are there any valid continuous Sudoku grids?
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 $[0,1]$.
Here's a weak solution. Using your favorite bijection, replace $[0,1]$ with the Cantor group $2^\mathbb N$ of infinite binary sequences. Then let $f(x,y)=x+y$. That is, just use the group operation: pointwise XOR.