Continuous vs discrete signal energy
I think you are correct. People are being fast and loose with the expression in your Eq. (2), but it captures the behaviour of the energy of the signal up to a constant $T_s$ factor (the sampling period), which is maybe why they do it. 

As you say, the signal energy is given by
$$
E_s = \int_{-\infty}^{+\infty} |x(t)|^2\;dt \tag{1}
$$
If you want to approximate this using a Riemann sum for the discrete case, the equivalent expression (depending on your choice of indexing) is
$$
E_s \approx \sum_{n=1}^{\infty} x[n]\; x^*[n]\cdot T_s \tag{2}
$$

You can see immediately that there must be a factor of $T_s$ in there otherwise the dimensions of the units aren't consistent. The correct units of signal energy as given by Eq. (1) are $[\textrm{signal}^2\cdot\textrm{time}]$. If you drop the $T_s$ then the units of your Eq. (2) would just be $[\textrm{signal}^2]$, which are NOT the correct units for signal energy. See this related question for the units.

EDIT---------------------------------

The OP cited the wiki reference as the source of the confusion, specifically with the discrete version of Parseval's theorem. Parseval's theorem in continuous time says that the energy in the time domain and frequency domain are equal, and is written as

$$
\int_{-\infty}^{+\infty} |x(t)|^2\;dt = \int_{-\infty}^{+\infty} |X(f)|^2\;df \tag{3}
$$
where $x(t)$ is the signal, and $X(f)$ is its Fourier transform, defined by
$$
X(f) = \int_{-\infty}^{+\infty} x(t) e^{-2\pi i f t}\; dt. \tag{4}
$$
You can verify together with Eq. (4) that the units of both sides of Eq.(3) are $[\textrm{signal}^2\cdot\textrm{time}]$, as they should be for signal energy. So far so good.

You can get a discrete version of Parseval's theorem by discretising both sides to get
$$
\sum_{n=0}^{N-1} |x[n]|^2\cdot T_s = \sum_{k=0}^{N-1} |X[k]|^2\cdot \Delta f \tag{5}
$$
where $T_s$ is the sampling period, and $\Delta f$ is the separation of frequency bins (I have assumed that the length of the vector $X[k]$ and $x[n]$ are both equal to $N$ - this would be different if you had zero-padding, but remember that the extra zeroes would not contribute any extra signal energy). The Fourier transform $X[k]$ is the discrete version of $X(f)$, and is given by 
$$
\begin{align}
X[k] &amp;= \sum_{n=0}^{N-1} x[n] \exp \Big(  -2\pi i \frac{kn}{N} \Big) \cdot T_s \tag{6}\\
&amp;= X_{\textrm{DFT}}[k]\cdot T_s \tag{7}
\end{align}
$$
and we recognise $X_{\textrm{DFT}}[k]$ as being the result of the discrete Fourier transform (DFT) operation - this you might obtain directly from the output of the FFT function, for example in MATLAB. There is an extra factor of $T_s$ in the expression for $X[k]$, because it is an approximation of the continuous Fourier transform from Eq. (4) - again needed for the units to add up.

Now we can insert Eq. (7) into Eq. (5), and use the fact that $\Delta f = F_s/N = 1/(NT_s)$, where $F_s$ is the sampling frequency, to obtain
$$
\begin{align}
\sum_{n=0}^{N-1} |x[n]|^2\cdot T_s &amp;= \sum_{k=0}^{N-1} |X_{\textrm{DFT}}[k]\cdot T_s|^2\cdot \Delta f \tag{8}\\
&amp;= \sum_{k=0}^{N-1} |X_{\textrm{DFT}}[k]|^2\cdot \frac{T_s^2}{NT_s}\tag{9}\\
\implies \sum_{n=0}^{N-1} |x[n]|^2 &amp;= \frac{1}{N}\sum_{k=0}^{N-1} |X_{\textrm{DFT}}[k]|^2\;,\tag{10}
\end{align}
$$
where the factors of $T_s$ have cancelled. This is the expression for Parseval's theorem given in the wikipedia link you cited, and is still valid. However, since we killed the factors of $T_s$, the units are now not signal energy units, and I personally wouldn't call the left hand side of Eq. (10) signal energy anymore (actually, it has units of signal power instead).

Hope that helps clear up a bit of the confusion. I always find it helpful to check the units at each stage as you go along, to keep track of things.