Both results are correct. Hence, the equality
$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=0}^{\infty}e^{-nT(1+j\omega)}\tag{1}$$
holds true.
One way to prove this is to realize that the term on the left-hand side of Eq. $(1)$ is periodic with period $2\pi/T$. Consequently, we can express this term by its Fourier series
$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=-\infty}^{\infty}c_ne^{jnT\omega}\tag{2}$$
It can be shown (try it!) that the Fourier coefficients $c_n$ are just (scaled) samples of the time domain function $x(t)$, so you obtain the right-hand side of Eq. $(1)$.
Eq. $(1)$ is an example of a more general result called Poisson's sum formula. The general form of Poisson's sum formula that specializes to Eq. $(1)$ is
$$\frac{1}{T}\sum_{k=-\infty}^{\infty}X\left(\omega-k\frac{2\pi}{T}\right)=\sum_{n=-\infty}^{\infty}x(nT)e^{-jnT\omega}\tag{3}$$
where $X(\omega)$ is the Fourier transform of $x(t)$.
Note that the right-hand side of Eq. $(3)$ is the discrete-time Fourier transform (DTFT) of the sequence $x(nT)$, which, according to Eq. $(3)$, equals the periodized spectrum $X(\omega)$ of the continuous-time function $x(t)$.
The dual form of Poisson's sum formula is
$$\sum_{k=-\infty}^{\infty}x(t-kT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}X\left(n\frac{2\pi}{T}\right)e^{jn\frac{2\pi}{T}t}\tag{4}$$
i.e., the Fourier coefficients of a periodized time domain function $x(t)$ are samples of its Fourier transform $X(\omega)$.
Note that choosing $\omega=0$ in $(3)$, or, equivalently, $t=0$ in $(4)$ leads to the remarkable result
$$\sum_{k=-\infty}^{\infty}x(kT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}X\left(n\frac{2\pi}{T}\right)\tag{5}$$