Is there a convolution mistake in my method?
You just multiplied the two functions and integrated them but you didn't convolve them. You must compute

$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau=2\int_0^2h(t-\tau)d\tau-\int_2^3h(t-\tau)d\tau$$

Alternatively, you can compute the step response

$$a(t)=\int_{-\infty}^th(\tau)d\tau$$

Because $x(t)=2u(t)-3u(t-2)+u(t-3)$ the output can be written in terms of the step response:

$$y(t)=2a(t)-3a(t-2)+a(t-3)$$