Where did we get the DC term of the Accumulator from DTFT?
Note that the equation
$$x[n]=y[n]-y[n-1]\tag{1}$$
doesn't uniquely determine the sequence $y[n]$. If some $y[n]$ satisfies $(1)$, so does $y[n]+c$ with some constant $c$. Consequently, Eq. $(1)$ determines $y[n]$ only up to a constant, which corresponds to a DC term in the frequency domain.
Hence, the equation
$$Y(e^{j\omega})=\frac{X(e^{j\omega})}{1-e^{-j\omega}}\tag{2}$$
is only valid for $\omega\neq 0$, or, more generally, for $\omega\neq 2\pi k$, $k\in\mathbb{Z}$.
The most straightforward way to derive the DTFT of
$$y[n]=\sum_{m=-\infty}^nx[m]\tag{3}$$
is to realize that $y[n]$ is the convolution of $x[n]$ with the unit step sequence $u[n]$, and, consequently, the DTFT of $y[n]$ is given by
$$Y(e^{j\omega})=X(e^{j\omega})U(e^{j\omega})\tag{4}$$
where $U(e^{j\omega})$ is the DTFT of the unit step $u[n]$:
$$U(e^{j\omega})=\pi\delta(\omega)+\frac{1}{1-e^{-j\omega}}\tag{5}$$
Several ways to derive the result $(5)$ are discussed in the answers to this and this question.
Note that in $(5)$ I use $\delta(\omega)$ instead of $\sum_k\delta(\omega-2\pi k)$ to avoid cluttered notation. The fact that the DTFT is always $2\pi$-periodic is understood.