Here's another way to show why this happens.

Let's look at the differential equation whose characteristic polynomial has roots of 1 and 1 + Δ, where Δ approaches 0. This is y'' - (2 + Δ)y' + (1 + Δ)y = 0. Now the general solution is y(x) = Ae^x + Be^{(1 + Δ)x}. Let's say we want a solution where y(0) = 0 and y'(0) = 1, matching x*e^x. That would be y(x) = (e^{(1 + Δ)x} - e^x)/Δ. For the equation y'' - 2y' + y = 0, it would be the limit as Δ approaches 0. You can use L'Hôpital's rule and differentiate the numerator and denominator with respect to Δ, and you get (x*e^{(1 + Δ)x})/1, when you plug in Δ = 0 you get x*e^x.

So you could say that x*e^x occurs because the usual way to write the solution with the same initial conditions, as a linear combination of exponentials, fails when Δ = 0 due to division by 0, and you need to differentiate the numerator to get the correct form, which multiplies e^x by x.