A Markov process with right continuous trajectories and left limits 
Let 
  $$
Nf(x) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-\frac{|x-y|^2}{2}}f(y)dy, \;\; f \in b\mathcal{B}(\mathbb{R}), x \in \mathbb{R}.
$$
  Let $X = (X_t, \mathcal{F}_t, \mathbb{P}^x)$ be a pure jump Markov process, with $(e^{tN})_{t \geq 0}$ as transition function. 
Show that $X$ is a $cadlag$ process.

Can someone please give me a hint on how to show that X is $cadlag$?

Your formula for the infinitesimal generator of $X$ needs to be adjusted slightly. It should be $Nf(x) ={1\over \sqrt{2\pi}}\int_{\Bbb R} e^{-(x-y)^2/2}[f(y)-f(x)]\,dy$.
The most direct way to proceed is to construct a cadlag process with $(e^{tN})_{t\ge 0}$ as transition semigroup. The process will be a compound Poisson process with the standard normal distribution as jump distribution. Let $(\Pi(t))_{t\ge 0}$ be a unit-rate Poisson process on $[0,\infty)$; that is, $\Pi(t)$ counts the number of arrivals in the interval $(0,t]$. Consequently, $t\mapsto\Pi(t)$ is cadlag. Next, let $(Z_n)_{n\ge 1}$ be an i.i.d. sequence of standard normal random variables, independent of $\Pi$. Finally, define
$$
X_t:=x+\sum_{n=1}^{\Pi(t)}Z_n,\qquad t\ge 0.
$$
Observe that $X$ is cadlag, and that $X$ jumps by a random (normal) amount each time $\Pi$ jumps up by $1$. You can show (by conditioning on the value of $\Pi(t)$) that $\Bbb E^x[f(X_t)] = e^{tN}f(x)$.
