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][1] that looks natural constructively, and can be considered a contrapositive of the usual [statement][2]) 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.

  [1]: https://mathoverflow.net/a/163965/44143
  [2]: https://en.m.wikipedia.org/wiki/Schanuel's_conjecture