As you guessed in the comments, it does not exist. The idea is the following: you are searching for some values that have small distance from some given values (the one on rectangles). If for example a new value $v$ must be close to some given values $v_1,v_2$, then necessarily $v_1, v_2$ must be close.
Conditions of Delta additivity in each variable is not enough to guarantee that $v_1$ and $v_2$, will be always close. If you want to find a condition, I guess you should do something like you do in elimination theory with equations, but in this "distance" fashion.
An interesting question so would be, mimicking elimination style: if such inequalities are satisfied, does there exist a solution? Intuitively I would say that if one among $f,g$ is actually a measure, then these inequalities are satisfied.
Without such conditions there is a counterexample. Take both sets to have 2 elements that we call $x,y$.
In the first one:
$x$ and $y$ have measure 0;
$\{x,y\}$ has measure 1.
In the second one:
$x$ and $y$ have measure $r$;
$\{x,y\}$ has measure $2r$.
Emptyset has zero measure in both. Note that singletons in the product has measure zero.
Now take the L-shaped set $L=\{(x,x), (x,y), (y,x) \}$ and suppose it has measure $A$. If we add the last brick to get the rectangle, we have $$ | A + 0- 2r| \le 1$$
If we take out the the brick $(y,x)$ the rectangle we are left with has projection $x$ on the first set, thus it has measure zero. On balance we get $$ |A -0-0| \le 1$$
In contradiction with the previous one for big $r$. Also, note that this yield that in general it does not exist a delta additive tensor product for any fixed $\delta$, and that even if one of them is a measure the tensor could not exist.