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removed some repetitive text and clarified the version of Schanuel’s conjecture being used
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I think the third condition is true conditional on Schanuel’s conjecture.

For example, we can use Schanuel’s conjecture three times to rule out the possibility that $e$ is the solution to $2^{2^u}-u=93$. I hope and expect that this (or some simplification) provides a template for ruling out all possible existential definitions of $e$.

If the above equation holds with $u=e$, then consider the four pairs of expressions:

$$w=\ln 2, \ e^w=2\\ x=1, \ e^x=e\\ y=e \ln 2, \ e^y=2^e\\ z=(2^e) \ln 2,\ e^z=2^{2^e}$$

We have five independent algebraic relationships among them:

$$e^z - e^x = 93\\ e^w = 2\\ x = 1\\ y = w e^x\\ z = w e^y$$

By Schanuel’s conjecture (in a version that looks natural constructively, and can be considered a contrapositive of the usual statement) there must be a rational linear dependence among the first four variables:

$$a w + b x + c y + d z = 0$$

So the three pairs of expressions with $w,x,y$ have four independent algebraic relationships between them, where the first comes from dividing the last equation by $\ln 2$:

$$a + b/w + c e^x + d e^y = 0\\ e^w = 2\\ x = 1\\ y = w e^x$$

By Schanuel’s conjecture there must be a rational linear dependence:

$$p w + q x + r y = 0$$

So the two pairs of expressions with $w,x$ have three independent algebraic relationships, where the first comes from dividing the last equation by $\ln 2$:

$$p + q/w + r e^x = 0\\ e^w = 2\\ x = 1$$

By Schanuel’s conjecture there must be a rational linear dependence

$$s w + t x = 0$$

This would mean $\ln 2$ is rational, which is impossible, so the original equation does not hold for $u=e$.

user44143