Let the Fourier transform of $f\in L^1(\Bbb R)$, denoted by $\mathcal{F}f$, be defined as
$$ \mathcal{F}f(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ixy} f(x)\,dx.$$
An oft-quoted result is that if $f,f'\in L^1(\Bbb R)$, then
$$ \mathcal{F}(f')(y) = iy\mathcal{F}f(y).\tag{1}$$
This is justified with integration by parts:
$$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ixy}f'(x)\,dx = \left.\frac{1}{\sqrt{2\pi}} e^{-ixy}f(x)\right|_{-\infty}^{\infty} +iy\int_{-\infty}^{\infty} e^{-ixy}f(x)\,dx.\tag{2}$$
The association between $(1)$ and $(2)$ is predicated on $f$ going to zero at $\pm\infty$, however there are plenty of differentiable $L^1$ functions which do not go to zero at infinity, e.g. take a compactly supported smooth function with support of width $\frac{1}{4}$, then center dilated copies of it at the integers with decreasing width away from zero. You could find a sequence of function values which never goes to zero in this case.
If $f$ were absolutely continuous, then it seems to me that this should be true since we would have (for any fixed $a$):
$$ f(x) = \int_a^x f'(y)\,dy + f(a).$$
Taking a limit as $x$ tends to infinity would give that
$$ f(\infty) = \int_a^{\infty} f'(y)\,dy + f(a).$$
Each quantity on the right hand side is well-defined and so $f(\infty)$ is a constant, but the only constant it could be (to retain integrability) would be zero.
However without assuming absolute continuity this seems no-go since if $f$ was the Cantor function, $f' \equiv 0$ almost everywhere so the left hand size of $(2)$ is zero, the first term on the right hand side is zero since the Cantor function is compactly supported and the second term on the right hand side is nonzero since the Fourier transform is injective. (Unless of course I am mistaken about this.)
What exactly are the right conditions for the association between $(1)$ and $(2)$ to hold (or where has my logic broken down)?
Your objection is indeed correct
Here is another demonstration:
$$\int_{\Bbb R} e^{-ixy}f'(x) dx = \int_{\Bbb R} \lim_{h\to 0} e^{-ixy}\frac{f(x+h)-f(x)}{h} dx $$
$$= \lim_{h\to 0} \frac{1}{h}\int_{\Bbb R} e^{-ixy} (f(x+h)-f(x))dx $$
$$= \lim_{h\to 0} \frac{1}{h}\int_{\Bbb R} e^{-i(u-h)y}f(u)du - \hat{f}(y)$$
$$= \lim_{h\to 0} \frac{e^{ihy}-1}{h} \hat{f}(y) = iy \hat{f}(y)$$
The key point is the interversion of the integral and limit, I let you choose your favorite theorem