For $x>0$, the event $\{X>x\}$ has probability $P\{X>x\}=1-F_X(x)=e^{-\lambda x} > 0$. Hence,
$$
\newcommand{\E}{\mathbb{E}} \E[X\mid X> x] = \frac{\E[X\,I_{\{X>x\}}]}{P\{X>x\}} \, ,
$$
but
$$
\E[X\,I_{\{X>x\}}] = \int_x^\infty t\,\lambda\,e^{-\lambda t}\,dt = (*)
$$
(using Feynman's trick, vindicated by the Dominated Convergence Theorem, because it is fun)
$$
(*) = -\lambda \int_x^\infty \frac{d}{d\lambda}(e^{-\lambda t}) \,dt = -\lambda \frac{d}{d\lambda} \int_x^\infty e^{-\lambda t} \,dt
$$
$$
= -\lambda \frac{d}{d\lambda} \left(\frac{1}{\lambda} \int_x^\infty \lambda\,e^{-\lambda t} \,dt\right) = -\lambda\frac{d}{d\lambda}\left(\frac{1}{\lambda}\left(1 - F_X(x)\right)\right)
$$
$$
= -\lambda\frac{d}{d\lambda}\left(\frac{e^{-\lambda x}}{\lambda}\right) = \left(\frac{1}{\lambda}+x\right)e^{-\lambda x} \, ,
$$
which gives the desired result
$$
\E[X\mid X>x] = \frac{1}{\lambda}+x = \E[X] + x \, .
$$