Ok, I was wrong obviously, PerManne's result is correct it took me a long time and quite a bit of paper to finally re-do everything.
For the Dirac part it is actually easier than I thought, just pick any function nice enough $g(x_1, x_2)$ and try to compute $I(t) = \int_{x_1}\int_{x_2}g(x_1, x_2)F(t, x_1, x_2)dx_1 dx_2$ where $F$ is defined as above, when $t \to T$.
Simple calculations lead to
$I(t)=\frac{1}{\sigma^2} \int_{x_1}\int_{x_2} \frac{g(x_1, x_2)}{\sqrt{2\pi \sigma^2 t}} e^{-\frac{(x_1-x_0)^2+(x_2-x_0)^2}{2\sigma^2(T+t)}} \frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}}dx_2dx_1$ $= \frac{1}{\sigma^2} \int_{x_1}\int_{x_2} h(t, x_1, x_2) \frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}}dx_2dx_1$
with $\hat{\sigma} = \sigma \sqrt{\frac{T^2-t^2}{t}}$, $\hat{\sigma} \to 0$ when $t \to T$.
Then the Dirac term comes from the second exponential since $\frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}} \to +\infty$ when $x_1 = x_2$ and $t \to T$ however $\int_{x_2}\frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}}dx_2 = 1$
So finally we get when $t \to T$, $I(T) \to \frac{1}{\sigma^2}\int_{x_1} h(x_1, x_1, T)dx_1$ with $h(x_1, x_1, T) = g(x_1, x_1)P(x_1, T \mid x_0, 0)$
(there might be a little more to say to prove this but this works)
And I guess we can write it like $I(T) \to \frac{1}{\sigma^2}\int_{x_1}\int_{x_2} g(x_1, x_2)P(x_1, T \mid x_0, 0) \delta(x_2-x_1) dx_ 1dx_2$ which is, without the integrals and the function $g$, what I asked.
Thanks a lot PerManne!