Sure, $f(x) \approx f(x-\delta x)$, so you can pair off most of the terms in pairs that are approximately $0$. But they are only approximately zero, and there are a lot of them ($(b-a)/\delta x$ of them, and $\delta x$ is very small!). So, the total sum of them may still be non-negligible because there are so many of them.
Indeed, using the same reasoning, you could just say that in $\sum f(x)\delta x$ each term is very small, since $f(x)$ is bounded and $\delta x$ is small. Your approach would then say that $0$ is a good approximation to the integral. Of course this is wrong, because although each term is small (roughly proportional to $\delta x$), the number of terms is large in a way that cancels it out (proportional to $1/\delta x$).
We can be more precise and say that $f(x)-f(x-\delta x)\approx f'(x) \delta x$ if $f$ is differentiable. So, the errors in your terms are also roughly proportional to $\delta x$, at least as long as $f'(x)$ is bounded and stays away from $0$. We can thus expect a non-negligible total error when we add up $(b-a)/\delta x$ such errors.
To be even more precise, we can expect the error to be $\int_a^b xf'(x)\, dx$, since each term of the error has the form $(x-\delta x)(f(x)-f(\delta x))\approx xf'(x)\delta x$. We can verify this rigorously: if you integrate $\int_a^b f(x)\, dx$ by parts with $u=f(x)$ and $v=x$ (assuming $f$ is continuously differentiable) you get $$\int_a^b f(x)\,dx=xf(x)\bigg|_a^b-\int_a^b xf'(x)\,dx=(bf(b)-af(a))-\int_a^b xf'(x)\,dx.$$ The first term is exactly what you find with your approximation, so the error is indeed $\int_a^b xf'(x)\,dx$. In this sense, then, the error you are making is akin to saying that the derivative of $xf(x)$ should be $f(x)$, which is what you get by differentiating just the $x$ factor but ignoring the fact that $f(x)$ may also be changing.
