"It turns out that not only is the exponential distribution memoryless, but it is also the unique distribution possessing this property." - Sheldon Ross, author of "A First Course in Probability (eighth edition)". The proof presented in the book isn't particularly rigorous (nor is it ment to be) so with some help from the the response of Michael Hardy in Uniqueness of memoryless property I believe I managed to arrive to the correct conclusion but I am uncertain and thus I seek some clarification regarding my result.
Ultimately the problem boils down to solving the equation $f(s+t)=f(s)f(t)$ where $f$ is non-negative (since it's a cumulative distribution function) and is evaluated at real non-negative values. I believe I managed to show that only decaying exponential functions of the type $\alpha^{-\lambda x}$, $\lambda>0$ could serve as solutions to the problem but Sheldon Ross claims that $e^{-\lambda x}$ is the only unique solution. This leads to me believe that perhaps I have done something wrong. Is he correct or incorrect? I apologize for not providing you with the proof I came up with, I am bit too lazy to write it down.
Edit: Book pdf, page $211$ if anyone is interested, perhaps I did not mention something important that might be in there.
first, the exponential distribution is not the only memoryless distribution: the geometric distribution, defined as $$ \mathbb P(X=n) = (1-p)^{n-1}\cdot p, \quad n \in \mathbb{N}, p \in (0,1), $$ is also memoryless.
What the author probably meant is that the exponential distribution is the only continuous (i.e absolutely continuous with respect to the Lebesgue measure) distribution with this property. If you want to prove this, you use the fact that exponential functions of the form $a^{\lambda x}$ (for different $a$, and $\lambda$) are the only continuous functions which satisfy $$ f(s+t)= f(s) + f(t), \quad \forall s,t. $$ Of course you can choose $a=e$ but this is not necessary. However for every $a$ and $lambda$ there exists a $\kappa$ such that $$ a^{\lambda x}=e^{\kappa x}, \quad x \in \mathbb{R}, $$ namely, $\kappa= \lambda \cdot \ln(a)$.