Charcoal, 23 bytes

≔⁰θＦ⊖φ≔÷⁺Ｘχ⁴⁹⁹θ⁻φιθ2.Ｉθ

Try it online! Link is to verbose version of code. Explanation:

≔⁰θ

Start with a running total of 0.

Ｆ⊖φ≔÷⁺Ｘχ⁴⁹⁹θ⁻φιθ

For each integer from 1000 down to 2, add 10**499 to the running total, then divide by that integer. This is equivalent to summing the reciprocals of the factorials from 2 to 1000 but without any loss of precision.

2.Ｉθ

Output e to 500 digits.

Charcoal, 23 bytes

≔⁰θＦ⊖φ≔÷⁺Ｘχ⁴⁹⁹θ⁻φιθ2.Ｉθ

Try it online! Link is to verbose version of code. Explanation:

≔⁰θ

Start with a running total of 0.

Ｆ⊖φ≔÷⁺Ｘχ⁴⁹⁹θ⁻φιθ

For each integer from 1000 down to 2, add 10**499 to the running total, then divide by that integer. This is equivalent to summing the reciprocals of the factorials from 2 to 1000 but without any loss of precision.

2.Ｉθ

Output e to 500 digits.

20 bytes to output without the decimal point:

≔⁰θＦφ≔⁺Ｘχ⁴⁹⁹÷θ⁻φιθＩθ

Try it online! Link is to verbose version of code. Explanation: As above but counts down to 1 and also adds the 10**499 after dividing so that the result includes the 0! and 1! terms.