Which of the following is true for $$y(x)$$ that satisfies the differential equation $$\frac{dy}{dx} = xy - 1 + x - y$$; $$y(0) = 0$$:

We have $$\dfrac{dy}{dx} = xy - 1 + x - y$$. Let us rearrange the right-hand side:

$$xy - y + x - 1 = y(x - 1) + (x - 1) = (x - 1)(y + 1)$$

So the differential equation becomes $$\dfrac{dy}{dx} = (x - 1)(y + 1)$$.

Separating variables: $$\dfrac{dy}{y + 1} = (x - 1)\,dx$$

Integrating both sides: $$\ln|y + 1| = \dfrac{x^2}{2} - x + C$$

Using the initial condition $$y(0) = 0$$: $$\ln|0 + 1| = 0 - 0 + C$$, so $$C = 0$$.

Therefore, $$\ln(y + 1) = \dfrac{x^2}{2} - x$$, which gives $$y + 1 = e^{x^2/2 - x}$$, i.e., $$y = e^{x^2/2 - x} - 1$$.

At $$x = 1$$: $$y(1) = e^{1/2 - 1} - 1 = e^{-1/2} - 1$$.

Hence, the correct answer is Option A.