This is actually a nasty business when one uses the Fourier transform of such things as a constant function on $\mathbf{R}$.
If you want to prove $u'=\delta$ in the distribution sense, you should use the correct way of calculating distribution derivatives as stated by L. Schwartz.
Write $\left\langle f,\varphi\right\rangle$ the value of the distribution $f$ applied to a test function $\varphi\in\mathcal{C}^\infty$ with a compact support. You have
$$\left\langle\delta,\varphi\right\rangle=\varphi(0).\tag 1$$
If the distribution $f$ is a locally integrable function then
$$\left\langle f,\varphi\right\rangle=\int_{\mathbf{R}}f(x)\varphi(x)\,\mathrm{d}x.$$
Note that $\delta$ is not such a function, so the above formula does not apply for $\delta$.
According to your notations, one can compute $\left\langle u,\varphi\right\rangle$:
$$\left\langle u,\varphi\right\rangle=\int_{\mathbf{R}}u(x)\varphi(x)\mathrm{d}x=\int_0^\infty\varphi(x)\mathrm{d}x.$$
To compute $u'$, apply now the definition of the derivative $f'$
$$\left\langle f',\varphi\right\rangle=-\left\langle f,\varphi'\right\rangle$$
to $f=u$. You get
$$\left\langle u',\varphi\right\rangle=-\left\langle u,\varphi'\right\rangle=-\int_0^\infty\varphi'(x)\mathrm{d}x.$$
As $\varphi$ has a compact support, it is equal to $0$ at infinity and the above equation gives
$$\left\langle u',\varphi\right\rangle=-\lim_{x\to\infty}\varphi(x)+\varphi(0)=\varphi(0)$$
which is exactly the definition of the distribution $\delta$ in equation (1). We just have proved that $u'=\delta$.
I am going to look at the solution to a general problem and then approach your problem.
$$ L(u) = f(x) \tag{1} $$
defined on $ a < x < b$ subject to homogeneous boundary conditions where $L$ is a Sturm Liouville operator
$$ L(u) \equiv \frac{\rm d }{\rm dx}\bigg( p \frac{\rm d }{\rm d x} \bigg) + q \tag{2} $$
now in the simple case we have $ p= 1$ and $ q=0$ and our operator is $ L= \frac{\rm d^{2} }{\rm dx^{2}}$ this can be solved by variation of paramters
$$ u = u_{1} v_{1} + u_{2}v_{2} \tag{3}$$
Now consider this problem
$$ \frac{\rm d^{2} u}{\rm dx^{2}} = f(x) \tag{4}$$
with boundary conditions
$$ BC1: u(0) = 0 \\BC2: u(L) = 0 \tag{5} $$
the homogeneous solutions correspond to
$$ u_{1}(x) = x \\ u_{2}(x) = L - x \tag{6}$$
when you integrate
$$ v_{1}(x) = \frac{1}{L} \int_{0}^{x} f(\xi) (L-\xi) d \xi \tag{7} $$
$$ v_{2}(x) = -\frac{1}{L} \int_{0}^{x} f(\xi) \xi d \xi + c_{2} \tag{8} $$
then the solution to the non-homogeneous boundary problem becomes
$$ u(x) = \frac{-x}{L} \int_{x}^{L} f(\xi)(L-\xi) d \xi -\frac{L-x}{L}\int_{0}^{x}f(\xi) \xi d\xi \tag{9} $$
we transform this into our equation
$$ u(x) =\int_{0}^{L}f(\xi) G(x|\xi) d \xi \tag{10}$$
where our green's function is
$$ G(x|\xi) =\begin{align}\begin{cases} c_{1}+c_{2}x & 0 < \xi \\ \\ d_{1}+d_{2}x & x > \xi \end{cases} \end{align} \tag{11}$$
applying the boundary conditions $G(0|\xi) =G(1|\xi) = 0$
$$ G(x|\xi) =\begin{align}\begin{cases} cx & x < \xi \\ \\ d(x-1) & x > \xi \end{cases} \end{align} \tag{12}$$
in order for the greens function to be continuous
$$ c\xi = d(\xi-1) \implies d = c\frac{\xi}{\xi -1} \tag{13}$$
this is called the jump condition
$$ \frac{\rm d}{\rm dx} c \frac{\xi }{\xi -1}(x-1) \Big|_{x=\xi} - \frac{\rm d}{\rm dx}cx \Big|_{x =\xi} = 1 \tag{14}$$
$$c \frac{\xi }{\xi -1} - c = 1 \\ c= \xi-1\tag{15}$$
then the Green's function is
$$ G(x|\xi) =\begin{align}\begin{cases} (\xi-1)x & 0 \leq x \leq \xi \\ \\ \xi(x-1) & \xi \leq x \leq 1 \end{cases} \end{align} \tag{16}$$
$$ g(x|\xi) = (x-1)\int_{0}^{x} \xi f(\xi) d\xi + x \int_{x}^{1} (\xi-1) f(\xi) d\xi \tag{17}$$
Best Answer
Actually, the non rigorous definition of $\delta$ stipulates that $\int\limits_a^b\delta(x)\mathrm dx$ is $1$ if $a\lt0\lt b$ (what you recalled in your post) but also that it is $0$ if $a\leqslant b\lt0$ or if $0\lt a\leqslant b$.
Hence $H(x)=\int\limits_{-\infty}^x\delta(\xi)\mathrm d\xi$ should be $0$ if $x\lt0$ and $1$ if $x\gt0$.