On a related question I calculated the GCD of $I = (2 + 3i)$ and $J = (1 - i)$ to be $1$.
Now I know that $\mathbb{Z}[i]$ is a principal ideal domain. And I also know that the greatest common divisor of two elements $a, b$ in a P.I.D. generates the ideal $(a) + (b)$, yes?
So does that mean that $1$ in this case generates the ideal $I + J$? And hence that the ideal $I + J$ is actually equal to the ring $\mathbb{Z}[i]$ itself? Have I got all my assumptions correct here?
Yes, in this case $I+J = \mathbb{Z}[i]$ itself for the reasons you specified. You could also see this explicitly by finding $a = a_1 + a_2 i$ and $b = b_1 + b_2i$ in the ring such that $a(2 + 3i) + b(1-i) = u$ for some unit $u$ in the ring. Indeed, for $a = -1$ and $b = 3i$ you can see that
$$ -(2+3i) + 3i(1-i) = (-2-3i) + (3i+3) = 1 \in I+J. $$