I think the third condition is false 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 $u=e$.
Update: two further examples
Here are examples adapting this argument for equations with multiple variables, and for equations that don’t define $e$ uniquely.
Can we jointly define $e$ and $\pi$ via $$2\cdot 2^u-v=10,\ \ 2^v u = 24\ ?$$ If so, we take $$w =\ln 2,\ \ x=1,\ \ y=e \ln 2,\ \ z =\pi \ln 2$$ and the five algebraic relationships $$e^w=2,\ \ x=1,\ \ y=w e^x,\ \ 2e^y-z/w=10,\ \ e^z y/w=24$$ which force a rational linear dependence $$aw + bx + cy + dz = 0$$ Therefore $w,x,y,e^w,e^x,e^y$ satisfy the four algebraic relationships $$e^w=2,\ \ x=1,\ \ y=w e^x,\ \ 2e^y-(aw+bx+cy)/(-dw)=10$$ which force a rational linear dependence $$pw+qx+ry=0$$ which is impossible as before.
Can we define $e$ via $$2^{u+3}-8 \cdot 2^u=0\ ?$$ We can take $$w =\ln 2,\ \ x=1,\ \ y=e \ln 2,\ \ z =(e+3)\ln 2$$ and the five independent algebraic relationships $$e^w=2,\ \ x=1,\ \ y=w e^x,\ \ 2e^z-8e^y=0,\ \ 3w + y = z$$ which indeed force the rational linear dependence $3w+y=z$, but this doesn’t give four independent algebraic relationships among $w,x,y,e^w,e^x,e^y$.
I hope either I or someone else will soon see how to explain this more systematically.