5 explicitly allow for negative $r_i$ so that the induction hypothesis works
$\mathbb P$ is countable. Moreover, any $f\in\mathbb P$ is analytic, hence it has only countably many zeros. Thus $\mathbb A_E$ is countable, and in particular, extra-transcendental reals exist, and $\mathbb R^+\smallsetminus\mathbb A_E$ has the power of continuum.
For a concrete example, the Lindemann–Weierstrass theorem implies that $e$ is extra-transcendental.
EDIT: To tie up a loose end, every nonzero $f\in\mathbb P$ has only finitely many positive real roots. Since $f(x)$ is eventually dominated by its nonzero term with the highest exponent, the roots are bounded. Similarly, $f(x)$ is dominated by the term with the smallest exponent when $x\to0+$, hence the roots are bounded away from $0$, i.e., they are contained in a compact subset of $(0,+\infty)$. However, choosing a branch of logarithm makes $f$ holomorphic in $U=\mathbb C\smallsetminus(-\infty,0]$, therefore it can have only finitely many roots in any compact subset of $U$.
In fact, if $f(x)=a_0x^{r_0}+a_1x^{r_1}+\cdots+a_nx^{r_n}$ (with $a_i$ nonzero)r_i\in\mathbb R$pairwise distinct,$a_i\in\mathbb R\smallsetminus\{0\}$), then$f$has at most$n$positive real roots. We can prove this by induction on$n$. If$n=0$, then$f(x)=a_0x^{r_0}$has no positive real root. Assume the statement holds for$n-1$, and without loss of generality let$r_0< r_1,\dots,r_n$. n-1$. Put $g(x)=a_0+a_1x^{r_1-r_0}+a_2x^{r_2-r_0}+\cdots+a_nx^{r_n-r_0}=f(x)/x^{r_0}$. Then every positive root of $f$ is also a root of $g$. Moreover, between each two consecutive roots of $g$, there is a root of its derivative $g'$. Since $g'\in\mathbb P$ g'$has at most$n$nonzero terms (the derivative of the constant$a_0$vanishes), it has at most$n-1$positive real roots by the induction hypothesis, thus$f$has at most$n$such roots. I guess that one could also prove a variant of Decartes' rule of signs for these generalized polynomials along similar lines. 4 number of zeros EDIT: To tie up a loose end, every nonzero$f\in\mathbb P$has only finitely many positive real roots. Since$f(x)$is eventually dominated by its nonzero term with the highest exponent, the roots are bounded. Similarly,$f(x)$is dominated by the term with the smallest exponent when$x\to0+$, hence the roots are bounded away from$0$, i.e., they are contained in a compact subset of$(0,+\infty)$. However, choosing a branch of logarithm makes$f$holomorphic in$U=\mathbb C\smallsetminus(-\infty,0]$, therefore it can have only finitely many roots in any compact subset of$U$. In fact, if$f(x)=a_0x^{r_0}+a_1x^{r_1}+\cdots+a_nx^{r_n}$(with$a_i$nonzero), then$f$has at most$n$positive real roots. We can prove this by induction on$n$. If$n=0$, then$f(x)=a_0x^{r_0}$has no positive real root. Assume the statement holds for$n-1$, and without loss of generality let$r_0< r_1,\dots,r_n$. Put$g(x)=a_0+a_1x^{r_1-r_0}+a_2x^{r_2-r_0}+\cdots+a_nx^{r_n-r_0}=f(x)/x^{r_0}$. Then every positive root of$f$is also a root of$g$. Moreover, between each two consecutive roots of$g$, there is a root of its derivative$g'$. Since$g'\in\mathbb P$has at most$n$nonzero terms (the derivative of the constant$a_0$vanishes), it has at most$n-1$positive real roots by the induction hypothesis, thus$f$has at most$n$such roots. I guess that one could also prove a variant of Decartes' rule of signs for these generalized polynomials along similar lines. 3 Lindemann–Weierstrass$\mathbb P$is countable. Moreover, any$f\in\mathbb P$is analytic, hence it has only countably many zeros. Thus$\mathbb A_E$is countable, and in particular, extra-transcendental reals exist, and$\mathbb R^+\smallsetminus\mathbb A_E$has the power of continuum. For a concrete example, the Lindemann–Weierstrass theorem implies that$e\$ is extra-transcendental.