Finding cumulative distribution function satisfying $ P(X \leq x \:\cup \: Y \leq y) = P(X+Y \leq x + y) $

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I'm trying to find a cumulative distribution function $ F $ for a distribution which satisfies the following property for two independent random variables $ X, Y \in [0, 1] $ :

$ P(X \leq x \:\cup \: Y \leq y) = P(X+Y \leq x + y) $

Using the independence of $ X $ and $ Y $, I'm looking to find a cumulative distribution function $ F $ which satisfies:

$ 1 - (1 - F(x)) \cdot (1-F(y)) = F(x + y) $

I already found out that I am basically looking for a solution for the above functional equation, however I have no experience in solving such equations. So any help towards a solution is highly appreciated.

Additionally I found out that the error is quite small for $ F(x) = x $. So I guess the shape of the correct cumulative distribution function is quite close to this one.

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There is no such $F$: if $f(x)=1-F(x)$ then we have $f(x+y)=f(x)f(y)$. Also $f$ is monotone. The only such functions are of the type $f(x)=e^{cx}$ so $F(x)=1-e^{cx}$. So your extra condition $F(1)=1$ can never be satisfied.