As I've mentioned before, trying to pass calculus through sheer memorization effort is doom to fail.
Here you are asking of a way to memorize the fact that the $e^{\ln 2} = 2.$ This is simply a consequence of what the logarithm function is supposed to be.
Logarithms are the inverses of exponentials. Inverse functions are functions that "undo" what the "original" function does. The logarithm base $a$ "undoes" exponentiation base $a$; exponentiation base $a$ "undoes" the logarithm base $a$. We define $\log_a(b)$ to be the number $r$ exactly when $a^r = b$. The statement "$\log_a(b)=r$" and the statement "$a^r = b$" mean the same thing, just like the statement "$\frac{6}{2}=3$" and the statement "$3\times 2= 6$" mean the same thing (because "$\frac{6}{2}$" means "the number that multiplied by $2$ gives $6$").
That is, we always have $ a^{\log_a(x)} = x\quad\text{and}\quad \log_a(a^x) = x\quad\text{for all values of }x$
These identities are simply a consequence of the fact that each function "undoes" the other function. It is a waste of mental effort to try to memorize them: there's just too many of them! One for every possible value of $a$. It's far better (and more productive in the long run) to understand why they are true, so that they can be internalized as conclusions rather than memorized "magic spells" that have no meaning.
In particular, since $\ln x$ is the logarithm base $e$ of $x$, $\ln x = \log_e(x)$, the fact that the exponential undoes the logarithm and the logarithm undoes the exponential means that $e^{\ln(x)} = x\quad\text{and}\quad \ln(e^x) = x\quad\text{for all values of }x.$
So $e^{\ln 5} = 5$, $\ln(e^3) = 3$, $e^{\ln\pi} = \pi$, $17^{\log_{17}(42.4)} = 42.4$, and so on and so forth.
Since we also have that, among the properties of logarithms, $r\ln(b) = \ln(b^r)$,
So: $\sinh(\ln(2)) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{e^{\ln 2} - e^{\ln(2^{-1})}}{2} = \frac{2 - 2^{-1}}{2} = \frac{\quad 2 - \frac{1}{2}\quad}{2} = \frac{2}{2} - \frac{\;\frac{1}{2}\;}{2} = 1 - \frac{1}{4} = \frac{3}{4}.$