How to show the difference of max and min of exponential random variables is exponential?

$M-L$ is exponentially distributed iff $\lambda_1 = \lambda_2$.

Claim: If $\lambda_1 = \lambda_2$, then $M-L \sim Expo(\lambda_1)$.

This is a standard result. Informally, if you think of $X, Y$ as the typical exponential waiting times for two different buses, then once the first bus arrived, $M-L$ is the time until the second bus arrives, but since the second bus is memoryless, it "didn't care" that time $L$ has elapsed. Hence $M-L \sim Expo(\lambda)$.

You can find more formal proofs, some using explicit integrals, in the links provided by StubbornAtom in the comments. Alternatively, I once wrote this proof which explicitly replicates the two-bus reasoning above. My proof does not involve evaluating integrals but is still rigorous (IMHO).

Claim: If $\lambda_1 \neq \lambda_2$, then $M-L$ is not exponential.

Again consider the two buses example. If the more frequent bus comes first, then $M-L \sim Expo(\min(\lambda_1, \lambda_2))$ because you must now wait for the rarer bus. If the rarer bus comes first, the $M-L \sim Expo(\max(\lambda_1, \lambda_2))$ because you are now waiting for the more frequent bus. So the overall distribution is a "mixture" and is not a single exponential.

E.g. both this MSE answer and these notes give the pdf $f_Z$ of $Z = X - Y$. Since $M-L = |X - Y|$, the pdf of $M-L$ is simply:

$$f_{M-L}(v) = f_Z(v) + f_Z(-v) = {\lambda_1 \lambda_2 \over \lambda_1 + \lambda_2} (e^{-\lambda_1 v} + e^{-\lambda_2 v}) ~~\text{for } v \ge 0$$

and of course $f_{M-L}(v) = 0$ for $v < 0$.