This answer ended up being much longer than I expected so tl/dr: The "invariances" Witten is talking about are redundancies in our description of the theory. So if we want to integrate over the physical space, we must integrate "up to diff and Weyl". If we did integrate over everything, we basically end up integrating over the same theory multiple times so we would be over-counting the integral.
Let's understand this by looking first at a much simpler example.
A circle $S^1$ can be described by an angle parameter $\theta \in [0,2\pi)$. Integrating a function $f(\theta)$ over the circle is simply
$$
I = \int_0^{2\pi} d\theta f(\theta).
$$
Alternatively, we can describe the circle as the entire real line $\theta \in {\mathbb R}$ with the identification $\theta \sim \theta + 2\pi$. In this description, a function $f(\theta)$, $\theta \in {\mathbb R}$ is a function on the circle only if $f(\theta+2\pi) = f(\theta)$. The periodicity in $f$ is required since $\theta$ is identified as shown previously. This description is expressed by $S^1 = {\mathbb R}/{\mathbb Z}$.
In this description integrating a function over the circle does not translate to integrating over the entire real line (again because of the identification in $\theta$). Rather, we must choose a "fundamental region" which represents the circle and we will integrate over this fundamental region. The fundamental region is chosen by requiring that every value of $\theta$ on the real line can be mapped to a value in the fundamental region using the identification for $\theta$. A particularly nice choice for the fundamental region is $[0,2\pi)$. It should be clear that every value of $\theta$ can be mapped to this region using the identification. For instance,
$$
\theta = 50.67 \sim 50.67-8(2\pi) = 0.4045 \in [0,2\pi).
$$
Integration over the circle is now the same thing as integrating over the fundamental region so we have
$$
I = \int_0^{2\pi} d\theta f(\theta).
$$
Note that we could have chosen ANY other fundamental region also and the result would have been the same. For instance, if we choose the region $[-2\pi,0)$, then
$$
I = \int_{-2\pi}^0 d\theta f(\theta) = \int_0^{2\pi} d\theta f(\theta-2\pi) = \int_0^{2\pi} d\theta f(\theta).
$$
In the last step, we have used the periodicity of $f$.
If we had integrated over the entire real line, we would have gotten a nonsensical answer, but one that has an interesting interpretation
$$
I' = \int_{\mathbb R} d\theta f(\theta) = \sum_{n=-\infty}^\infty \int_{2n\pi}^{2(n+1)\pi} d\theta f(\theta) = \sum_{n=-\infty}^\infty \int_0^{2\pi} f(\theta) = \sum_{n=-\infty}^\infty I
$$
Integrating over ${\mathbb R}$ is therefore the same as integrating over $S^1$ (infinitely) many times. $I'$ is clearly divergent due to the infinite sum, but we can rewrite this in an interesting way. The infinite sum is the volume of the group ${\mathbb Z}$ (i.e. the number of elements) so we can write
$$
I' = \text{vol}({\mathbb Z}) I \quad \implies \quad I = \frac{I'}{\text{vol}({\mathbb Z})}.
$$
Both the numerator and denominator are infinite, but the "infinities cancel" to give a finite answer at the end.
Let us go back to the original question posed by OP. Everything should become clear if we simply relate everything Witten said (or was meaning to say, IMHO) to the circle example presented above.
Gravity is the dynamical theory of metrics (metrics $\leftrightarrow \theta$). However, not all metrics are "different". Rather, any two metrics related by diffeomorphisms and Weyl transformations are identified,
$$
\text{metric} \sim \text{metric} + \text{diff and Weyl} \quad \leftrightarrow \quad \theta \sim \theta + 2\pi
$$
All physical functions must be "invariant" under these transformations
$$
f(\text{metric}) = f( \text{metric} + \text{diff and Weyl}) \quad \leftrightarrow \quad f(\theta) = f(\theta + 2\pi).
$$
In the path integral, our goal is to integrate over "different" metrics. But remember, not all metrics are different - some are identified. Thus, as in the circle example, we must choose a fundamental region and only integrate over that region. This is what Witten means when he says integrate "up to diff and Weyl".
Note that following the last circle example, there is a better way to describe such an integral. We could instead integrate over ALL metrics and then divide by the volume of diff and Weyl, i.e.
$$
I = \frac{I(\text{all metrics})}{\text{vol(diff and Weyl)}} \quad \leftrightarrow \quad I = \frac{I'}{\text{vol}({\mathbb Z})} .
$$
In practice, this is often the way integrals over metrics are performed. They are much easier to do since we don't have to this "up to diff and Weyl" business. As long as we divide by the volumes the answer is the same.
