Is my conjecture below true?
It's a new conjecture. It's an extension of Lin's theorem in [Lin 1983] which is cited in [Chow 1999] - see both references at the bottom of this page.
It seems that Ferng-Ching Lin was a student of Stephen Hoel Schanuel: https://www.genealogy.math.ndsu.nodak.edu/id.php?id=192925
The proof is very simple because the preconditions of my conjecture contain the preconditions of Lin's theorem except irreducibility. The irreducibility follows directly from very simple arguments from elementary algebra.
$\mathbb{L}$ are the elementary numbers, or Liouvillian numbers:
http://mathworld.wolfram.com/LiouvillianNumber.html, https://en.wikipedia.org/wiki/Elementary_number, http://mathworld.wolfram.com/ElementaryNumber.html, [Chow 1999].
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Theorem [Lin 1983]:
If Schanuel's conjecture is true and $\tilde{P}(X,Y)\in\overline{\mathbb{Q}}[X,Y]$ is an irreducible polynomial involving both $X$ and $Y$ and $\tilde{P}\left(\alpha,e^{\alpha}\right)=0$ for some non-zero $\alpha$ in $\mathbb{C}$, then $\alpha$ is not in $\mathbb{L}$.
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Conjecture:
Let
$f$ a non-constant elementary function,
$n\in\mathbb{N}_{\ge 1}$,
$P(x,y)\in\overline{\mathbb{Q}}[x,y]\setminus(\overline{\mathbb{Q}}[x]\cup\overline{\mathbb{Q}}[y])$ irreducible over $\mathbb{C}$,
$Q(x,y)\in\overline{\mathbb{Q}}[x,y]$ not zero,
$p(x)\in\overline{\mathbb{Q}}[x]$,
$q(x)\in\overline{\mathbb{Q}}[x]$ not zero,
$R(x,y)=\frac{P(x,y)}{Q(x,y)\cdot q(x)^n}$,
$r(x)=\frac{p(x)}{q(x)}$ not constant
so that
$P(x,y)$ and $Q(x,y)$ coprime over $\mathbb{C}$,
$p(x)$ and $q(x)$ coprime over $\mathbb{C}$.
Suppose Schanuel's conjecture is true.
If a $\tilde{P}(x,y)\in\overline{\mathbb{Q}}[x,y]\setminus(\overline{\mathbb{Q}}[x]\ \cup\ \overline{\mathbb{Q}}[y])$ of $degree_x=n$ with $\frac{P(x,y)}{{q(x)}^n}=\tilde{P}(r(x),y)$ exists and $R(f(z_0),e^{r(f(z_0))})=0$ for $z_0\in\mathbb{C}$ and $r(f(z_0))\neq 0$, then $z_0$ is not an elementary number.
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Proof draft:
We will prove our conjecture using Lin's theorem. Lin's theorem makes statements about equations of the form $\tilde{P}(\alpha,e^\alpha)=0$, wherein $\alpha\in\mathbb{C}$ and $\tilde{P}(x,y)$ is irreducible over $\mathbb{C}$. We consider the equation $R(f(z_0),e^{r(f(z_0))})=0$ from our conjecture and will show that under the preconditions of our conjecture there exists an equation equivalent to it in the form from Lin's theorem.
Setting $f(z_0)=x$ and $e^{r(f(z_0))}=y$ gives the equation $R(x, y)=0$. We consider it in the form $\frac{P(x, y)}{Q(x, y)\cdot q(x)^n}=0$ from the preconditions of our conjecture. Because $P(x,y)$ and $Q(x,y)$ are coprime over $\mathbb{C}$ by precondition, we can multiply both sides of the equation by $Q(x,y)$ and get the equation $\frac{P(x,y)}{q(x)^n}=0$ equivalent to it. By precondition, $\frac{P(x,y)}{{q(x)}^n}=\tilde{P}(r(x),y)$ holds. So our equation becomes $\tilde{P}(r(x),y)=0$. Backsubstitution of $x$ and $y$ gives $\tilde{P}(r(f(z_0)),e^r(f(z_0)))=0$.
We have $\forall z_0\in\mathbb{C}\colon r(f(z_0))\in\mathbb{C}$. We can therefore set $r(f(z_0))=\alpha$, which gives the form $\tilde{P}(\alpha,e^\alpha)=0$ from Lin's theorem. To fulfill all preconditions from Lin's theorem, we still have to show that $\tilde{P}(x,y)$ is irreducible over $\mathbb {C}$.
If $\tilde{P}(x,y)$ is reducible over $\mathbb{C}$, then
$n_1\in\mathbb{N}_{\ge 1}$,
$n_2\in\mathbb{N}_0$,
$a_0,...,a_{n_1}\in\mathbb{C}[y]$,
$b_0,...,b_{n_2}\in\mathbb{C}[y]$ not all constant
exist so that
$a_{n_1},b_{n_2}\neq 0$,
$\tilde{P}(x,y)=(a_0+a_1x^1+...+a_{n_1}x^{n_1}) \cdot(b_0+b_1x^1+...+b_{n_2}x^{n_2})$.
The highest degree regarding $x$ of the polynomial product on the right-hand side of the equation is $n_1+n_2$. This must be equal to the highest degree regarding $x$ of the polynomial $\tilde{P}(x,y)$ on the left-hand side of the equation. This was $n$ according to the preconditions of our conjecture. We have therefore $n_1+n_2=n$.
By substitution of $x$ by $r(x)$, we get
$$\tilde{P}(r(x),y))=(a_0+a_1r(x)^1+...+a_{n_1}r(x)^{n_1})\cdot(b_0+b_1r(x)^1+...+b_{n_2}r(x)^{n_2}).$$
Because, according to the preconditions of our conjecture, $r(x)=\frac{p(x)}{q(x)}$, and $p(x)$ and $q(x)$ are coprime over $\mathbb{C}$, we have
$$\tilde{P}(r(x),y))=(a_0+a_1\left(\frac{p(x)}{q(x)}\right)^1+...+a_{n_1}\left(\frac{p(x)}{q(x)}\right)^{n_1})\cdot(b_0+b_1\left(\frac{p(x)}{q(x)}\right)^1+...+b_{n_2}\left(\frac{p(x)}{q(x)}\right)^{n_2})$$
$$=(a_0+a_1\frac{p(x)^1}{q(x)^1}+...+a_{n_1}\frac{p(x)^{n_1}}{q(x)^{n_1}})\cdot(b_0+b_1\frac{p(x)^1}{q(x)^1}+...+b_{n_2}\frac{p(x)^{n_2}}{q(x)^{n_2}})$$
$$=(a_0+\frac{a_1p(x)^1}{q(x)^1}+...+\frac{a_{n_1}p(x)^{n_1}}{q(x)^{n_1}})\cdot(b_0+\frac{b_1p(x)^1}{q(x)^1}+...+\frac{b_{n_2}p(x)^{n_2}}{q(x)^{n_2}})$$
$$=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})}{q(x)^{n_1}}\cdot\frac{(b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^{n_2}}$$
$$=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^{n_1+n_2}}.$$
Because, as stated above, $n_1+n_2=n$, we have
$$\tilde{P}(r(x),y))=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$
According to the preconditions of our conjecture, $\tilde{P}(r(x),y))=\frac{P(x,y)}{q(x)^n}$. We have therefore
$$\frac{P(x,y)}{q(x)^n}=\frac{(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2})}{q(x)^n}.$$
Multiplying by $q(x)^n$ yields
$$P(x,y)=(a_0q(x)^{n_1}+a_1p(x)^1q(x)^{n_1-1}+...+a_{n_1}p(x)^{n_1})\cdot (b_0q(x)^{n_2}+b_1p(x)^1q(x)^{n_2-1}+...+b_{n_2}p(x)^{n_2}).$$
If $\tilde{P}(x,y)$ is reducible over $\mathbb{C}$, then $P(x,y)$ is accordingly reducible over $\mathbb{C}$. Because, according to the preconditions of our conjecture, $P(x,y)$ is irreduible over $\mathbb{C}$, $\tilde{P}(x,y)$ is irreducible over $\mathbb{C}$.
Thereby all preconditions from Lin's theorem are fulfilled and we can apply it to our equation. According to Lin's theorem, $\alpha$ is not in $\mathbb{L}$. Because $\alpha=r(f(z_0))$, $r(f(z_0))\notin\mathbb{L}$. Because $r$ and $f$ are elementary functions, $r(f(z))\in\mathbb{L}$ holds for all $z\colon z\in dom(r\circ f)\ \land\ z\in\mathbb{L}$. But because, as just stated, $r(f(z_0))\notin\mathbb{L}$, $z_0\notin\mathbb{L}$, so $z_0$ is not an elementary number.
q.e.d.
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[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
Here is a recent article using another method, Khovanskii's method:
[Belov-Kanel/Malistov/Zaytsev 2020] Belov-Kanel, A.; Malistov, A.; Zaytsev, R.: Solvability of equations in elementary functions. Journal of Knot Theory and Its Ramifications 29 (2020) (2) 204-205
