Is there an asymmetry between our ability to compute $e^x$ for rational $x$ vs for irrational $x$? Assume we know the value of the number $e$. Consider the meaning of the symbol
$$e^x$$
If $x$ is rational then

*

*$e^0$=1

*$e^n=\prod\limits_{i=1}^n e$, for $n\in\mathbb{N}$

*$e^{-n}=\frac{1}{e^n}=\frac{1}{\prod\limits_{i=1}^n e}$, for $n\in\mathbb{N}$

*$e^{1/n}=\sqrt[n]{e}$, for $n\in\mathbb{N}$

*$e^{m/n}=(\sqrt[n]{e})^m$ for $m,n\in\mathbb{N}$
So at least superficially it seems we know what $e^x$ is in terms of other defined operations such as multiplication, division,  and n-th root.
Now, if $x$ is irrational then $e^x$ is defined as
$$e^x=\exp(x)=\log^{-1}(x)$$
So to compute, for example $e^{\sqrt{2}}$ we would need to figure out what
$$\log^{-1}{\sqrt{2}}$$
is.
Is there an actual asymmetry between our ability to compute $e^x$ for rational $x$ vs for irrational $x$ or is this impression incorrect?
EDIT: I not looking for a way using series. I am following the book Calculus by Spivak. So far limits, continuity, derivatives, integrals, the fundamental theorem of calculus, trigonometric integrals, and exponential and logarithm functions have been covered. Series is still a few chapters aways.
 A: 
So at least superficially it seems we know what $e^x$ is in terms of other defined operations such as multiplication, division, and n-th root.

This isn't quite right; these identities don't help us "solve for $e$" they "depend on $e$".  In much the same way that the following identities "depend on $2$":
$$
2^0=1\\
2^n=\prod\limits_{i=1}^n 2 \quad \mbox{for $n\in\mathbb{N}$}\\
2^{-x}=\frac{1}{2^x}\\
2^{1/n}=(\sqrt[n]{2})    \quad \mbox{for $n\in\mathbb{N}$}\\
2^{m/n}=(\sqrt[n]{2})^m   \quad \mbox{for $n,m\in\mathbb{N}$}\\
$$
This is because the identity $a^{(x+y)}=a^x a^y$ is valid for any $a\in\mathbb{R}_{\geq0}$ and so you can derive both the $e^{(x+y)}=e^x e^y$ identity (which you mention in the comments) as well as a $2^{(x+y)}=2^x2^y$ identity (which is probably the simplest non-trivial example).

All of the identities you mention (and all of the ones I mention) can basically be derived from the basic $a^{(x+y)}=a^x a^y$ starting point.
If $x=x$ and $y=0$ then:
$$
2^{(x+0)}=2^{x}2^{0} \quad \rightarrow \quad 2^{0}=1
$$
If you consider $x=n-1$ and $y=1$, then you can see that there is essentially an repeatable "pull out $2^1$" process that you can apply until $n$ is fully sperated:
$$
2^{n}=2^{(n-1)}2^{1} \quad \rightarrow \quad 2^{n}=\underbrace{2^{1}2^{1}2^{1}...2^{1}}_{n}
$$
And the exact same sorts of derivations can be done starting from the $e^{(x+y)}=e^x e^y$ identity (in fact I'd be surprised if those derivations don't show up either in the textbook or its exercises).
Basically, your book is trying to get you comfortable manipulating equations that look like $e^{3+x}=e^{6x^2}$ in much the same way that you might already be able to handle something like $2^{3+x}=2^{6x^2}$.  In the end, that'll usually mean shuffling things around until you can finally get to something like $2^{3+x-6x^2}=2^{0}$... but where do you go from there?
Assuming this is a basic "solve for x" kind of problem we'd ideally like to keep simplifying the problem and getting x by itself.  And this is where Logarithms come into play as they can actually allow us to further simplify equations like the one above.  The Binary Logarithm (of x) is a function with the property:
$$ 
\log_2(2^x)=x
$$
which means that we can take our expression $2^{3+x-6x^2}=2^{0}$ and "take the binary logarithm of" both sides (just as we'd "subtract two" from both sides, or "multiply by y" both sides, or "exponentiate to the power 3" on both sides).
$$ 
\log_2(2^{3+x-6x^2})=\log_2(2^0) \quad \rightarrow \quad {3+x-6x^2} = 0
$$
from which you can rearrange and solve the quadratic equation for x.
The Natural Logarithm is just the $e$-version (compared to the above 2-version) that does the same kind of operation:
$$ 
\log_e(e^x)=x
$$
which we can write a few different ways according to that three-way definition at the end of your question:
$$ 
\log_e(e^x)=x\\
\log_e(\ \exp(x) \ )=x\\
\log_e( \ \log^{-1}_e(x) \ )=x\\
$$
Which is to say... that we have another tool (logarithms) that allow us to do a new kind of manipulation to equations beyond just the old add/subtract, multiply/divide, and exponentiation that we are more familiar with.
