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.

By removing a similar null set from $P$ we can also ensure it doesn't contain sets $\{re^{i\theta}:r\in E, \theta\in F\}$, where $E,F$ have positive measure (for example we can do the same construction locally in polar coordinates).

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.