Yes, we can do this by a small modification of the original argument, see the linked question. I'll describe the whole argument again here, but if you already read those answers, then the short version is that we remove successively less of each $I_n(p)$ as we approach the boundary of that interval. This allows us (for each $x\in I_n(p)$) to replace $I_n(p)$ by its maximal subinterval centered at $x$ while only slightly disturbing the approximate densities.
Construct open sets $G_1\supset G_2 \supset \ldots \supset E$, as follows. Start with any open $G_1\supset E$, and write $G_1 = \bigcup I_1(p)$ as the union of its components. For each $I_1(p)$, we construct an open set $G_2(p)$ with $E\subset G_2(p)\subset I_1(p)$. This $G_2(p)$ will cover only a small portion of $I_1(p)$ (possible since $|E|=0$), and, moreover, this portion will not be unduly concentrated near the boundary of $I_1(p)$ (this is the modification). More specifically, if $I_1(p)=(a,b)$, then we demand that
$$
|(a,a+h)\cap G_2(p)| \le 2^{-1}h \quad\quad\quad (1)
$$
for all $h>0$, and similarly near $b$. We can do this by cutting $I_1(p)$ into pieces $(a+2^{-k}, a+2^{-k+1})$ and treating these separately (if an endpoint is in $E$ here, I very slightly move such a point).
Then we put $G_2=\bigcup G_2(p)=\bigcup I_2(q)$. We repeat this whole procedure, with $2^{-1}$ replaced by $2^{-2}$ in (1), to obtain $G_3$ etc., and we finally set $F=\bigcup (G_{2n-1}\setminus G_{2n})$ (start with $G_1$, remove $G_2$, add $G_3$ etc. etc.).
If $x\in E$, then $x\in I_n(p_n)$ for some (unique) $p_n$ for all $n$. If $n$ is odd and we write $I_n(p_n)=(a,b)$ and $x$ is closer to $a$ than to $b$, then we consider $J=(a,a+2(x-a))$. Then $J\cap F\supset J\setminus G_{n+1}$, so (1) says that $|J\cap F|/|J|$ is almost $1$ if $n$ is large.
Similarly, we find intervals for which this ratio is almost zero from the even $n$'s.
Let us deal with each weakening one-by-one:
Dropping (i): In this case Lebesgue outer measure is a total extension of Lebesgue measure that satisfies all other requirements.
Dropping (ii): Already addressed in Gerald's answer - There are scaling invariant (property (iv)) Banach measures on $\mathbb{R}$. (There are also non scaling invariant Banach measures.)
Dropping (iii): There is no such total extension $\mu$. To see this choose a basis $B$ of the vector space $(\mathbb{R}, +)$ over the field $\mathbb{Q}$ of rationals with $1 \in B$. Let $W$ be the $\mathbb{Q}$-linear span of $B \setminus \{1\}$ and $X = W + 1$. Then the rational multiples of $X$ are pairwise disjoint and cover $\mathbb{R}$. By (i) + (ii) = countable additivity and (iv), we must have $\mu(X) > 0$ and hence, $\mu(X \cap [n, n+1)) > 0$ for some integer $n$. Let $Y = X \cap [n, n+1)$ and $Y_k = (1/k)Y$ for $k \geq 1$. Then $\{Y_k: k \geq 1\}$ is a family of pairwise disjoint sets whose union is bounded and the sum of whose $\mu$-measures is infinite (by (iv)): Contradiction.
Dropping (iv): Vitali's example shows that no total extension exists.
Other results along these lines:
(1) (Solovay, 1971) The existence of a total extension of Lebesgue measure that satisfies (i) + (ii) (= countable additivity) is equiconsistent with the existence of a measurable cardinal.
(2) (Ciesielski-Pelc, 1985) There is no maximal isomoetric-invariant extension of Lebesgue measure.
Best Answer
Firstly, a set $P$ of positive measure need not contain anything of the form $A\times B$, for example consider for some $k\in\mathbb{R}\setminus\{0\}$ the set $P=\{(x,y)\in\mathbb{R}^2;x-ky\not\in\mathbb{Q}\}$. Then the complement of $P$ is a null set, and any set $A\times B$ where $A,B$ have positive measure contains some point $(x,y)$ with $x-ky\in\mathbb{Q}$. This is because the set $A-kB$ contains an open interval. We can substitute $\mathbb{Q}$ by any other set dense in $\mathbb{R}$.
Now let $P=\{(x,y)\in\mathbb{R}^2;x\not\in\mathbb{Q},y\not\in\mathbb{Q},x-y\not\in\mathbb{Q}\}$. Any rotation/translation of $P$ cannot contain a set $A\times B$, with $A,B$ of positive measure, because then $A\times B$ would not intersect some set of parallel lines dense in $\mathbb{R}^2$ and with non zero slope, and we get a contradiction as in the previous case.
The same set $P$ also doesn't contain sets $\{re^{i\theta}:r\in E, \theta\in F\}$, where $E,F$ have positive measure. To check that, it will be enough to check that every point $p_0\in\mathbb{R}^2\setminus\{0\}$ with polar coordinates $(r_0,\theta_0)$ has a neighborhood $U$ such that $U$ doesn't contain sets $\{re^{i\theta}:r\in E, \theta\in F\}$, where $E,F$ have positive measure.
This is true because near $p_0$, any set of parallel lines dense in $\mathbb{R}^2$ can be expressed as $\{r(k\cos(\theta)+l\sin(\theta))\not\in Q\}$, for $k,l$ not both $0$ and $Q$ some dense set in $\mathbb{R}$. We can suppose that $p_0\not\in\{k\cos(\theta)+l\sin(\theta)=0\}$, so after changing $Q$ by $-Q$ if necessary, we can take logarithms and the equation of the parallel lines becomes $f(r)+g(\theta)\not\in\{\ln(q);q\in Q\cap\mathbb{R}^+\}$, where $f(r)=\ln(r),g(\theta)=\ln(k\cos(\theta)+l\sin(\theta))$. However, if $E,F$ have positive measure, then $f(E)$ and $g(F)$ have positive measure, so $f(E)+g(F)$ contains some open interval, so the same reasoning of before works.