If $S=\{a \in \mathbb{R^+}: a \ne1\}$ with binary operation * defined by $a*b=a^{\ln(b)}$. What is the inverse of $a$? To first solve for the identity, I did:
$a*e=a^{\ln(e)}=a$
$e*a=e^{\ln(a)}=a$
So in both cases, I'm unable to solve for $e$. Would this simply mean that there's no identity in the set? And so it follows that there's no way that any element in the set has an inverse?
 A: It means $e \approx 2.718 \ldots $ is the identity in this group as you have shown in your question.  I think you think $e$ is the symbol for identity, but it is Euler's number, the base of the natural logarithm.  This looks like the inverse of $a$ is $e^{\frac{1}{\ln(a)}}$
A: You are confusing "$e$" as denoting Euler's constant (which is relevant) and "$e$" as denoting the operative identity (which might be Euler's constant but we can not assume that without proving).  
Let's avoid the confusion.  Let's use,.... $z$ to mean the operative identity.
So for all $a$ then $a*z = a^{\ln z} = a$ for all $a$ and $z*a = z^{\ln a} = a$.
If $a \ne 1$ then $a^{m} = a \iff m = 1$ so $\ln z = 1$ so $z = $ Euler's constant.
And if $z^{\ln a} = a$ then $\ln (z^{\ln a}) = \ln a *\ln z = \ln a \implies \ln z = 1 \implies z = $ Euler's constant.
So the solution IS $z = e$.  
So now you must find the inverse of $a$.  $a*a^{-1} = a^{\ln a^{-1}} = e$.
For a given $a$, if we are told that $a^{\ln m} = e$ what is the value of $m$?
Well, take the natural log of both sides....
$\ln a^{\ln m} = \ln e$
$\ln m* \ln a = 1$
$\ln m = \frac 1{\ln a}$
$m = e^{\ln m} = e^{\frac 1{\ln a}}$.
So ... that's it.  $a^{-1} = e^{\frac 1{\ln a}}$.
A: Let  C be the identity element 
Then  $$\ {c} \ast {a} = {a} \ast {c} = {a} $$
 $$\ {a} \ast {c} = {a}^{lnc} = {a} $$
 Taking logrithm to both sides 
$$\ (lna)(lnc) = (lna) $$
So we get $$\ c = e (Euler's  number) $$
If you do the same with $$\ {c} \ast {a} $$
You will get the same answer. Therefore now we have c(identity element)= e
Now for the inverse we need to have $$ {a} \ast {b} = {c} $$ where 'b' is the inverse 
Then by doing a bunch of simple calculation you will get $$\ {b} = {e}^{(\frac{1} {lna})} $$
