For an Exponential distribution X with mean 500, we could say that P(X>1000 | X>500) = P(X>500) and the mean of the conditional distribution X|X>500 would also be 500. Good so far?
Can the memoryless aspect be extended in the other direction, so that you could say then that P(X<200 | X<400) = P(x<200)? It seems not, because the following problem's solution implied to me that the conditional distribution has a different mean.
The chore was to find the mean survival time of a component whose lifespan is exponentially distributed with mean 5, if the component survives less than 10 years. The answer was 3.44.
So to me this says that mean of the conditional distribution of X | X<10 is not the same as the mean of X.
One can think of exp distribution as waiting time for the first event in a Poisson process. The memoryless property simply says that waiting does not help - the probability of an event occurring in the next 5 minutes is the same, no matter how much you waited already (no matter how on what you condition, $X>500$ in your case). Hence, $P(X>500+500\vert 500)=\Pr(X>500)$.
Clearly, the opposite is not true. If you know you've waited for less than 500 than the event time is between 0 and 500, and the bounded support alone should hint that the distribution is not exponential.
If components last for less than 10 years and nine already passed, then your component will break very soon, and with probability 1 will break in the next year. This is not the probability of breaking in the first year of its life.