You're right about your intuition, but note that the differentiation property is always valid. This can be seen as follows:
If $x(t)$ is a real-valued function which is constant everywhere except at $t=0$ where it has a discontinuity such that $x(0^+)-x(0^-)=1$, then you have
$$\frac{dx(t)}{dt}=\delta(t)\tag{1}$$
In the frequency domain, this is equivalent to
$$j\omega X(j\omega)=1\tag{2}$$
From $(2)$ we can conclude that
$$X(j\omega)=\frac{1}{j\omega}+c\delta(\omega),\quad c\in\mathbb{R}\tag{3}$$
because $j\omega\delta(\omega)=0$.
Note that the real part of $X(j\omega)$ corresponds to the even part of $x(t)$, and the imaginary part corresponds to the odd part of $x(t)$. For the given $x(t)$, the even part is just its DC value. So if $x_0$ is the DC value of $x(t)$, i.e.,
$$x_0=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}x(t)dt\tag{4}$$
then we must have
$$\mathcal{F}\{x_0\}=2\pi x_0\delta(\omega)=c\delta(\omega)\tag{5}$$
from which we obtain
$$c=2\pi x_0\tag{6}$$
Consequently, the Fourier transform of $x(t)$ is given by
$$X(j\omega)=\frac{1}{j\omega}+2\pi x_0\delta(\omega)\tag{7}$$
For $x(t)=\frac12\textrm{sgn}(t)$ we have $x_0=0$ and consequently,
$$\mathcal{F}\left\{\frac12\textrm{sgn}(t)\right\}=\frac{1}{j\omega}\tag{8}$$
and for $x(t)=u(t)$ we have $x_0=\frac12$ and
$$\mathcal{F}\left\{u(t)\right\}=\frac{1}{j\omega}+\pi\delta(\omega)\tag{9}$$
Also take a look at this and this question, which deal with the same problem in discrete time.
