Answer
See explanation.
Work Step by Step
First, note that the cdf of $X$ is given by
$\begin{align*}
t\ge 0, F_X(t) &= P(X\le t) \\
&= \int_{-\infty}^t f_X(x)\ dx \\
&= \int_0^t (1/\lambda)e^{-x/\lambda}\ dx \qquad \biggl[\ \text{since}\ f_X(x)=0, x\lt 0\ \biggr]\\
&= -e^{-x/\lambda}\biggr\vert_0^t \\
&= -e^{-t/\lambda} - (- e^{-0}) \\
F_X(t) &= 1 - e^{-t/\lambda}, t\ge 0.
\end{align*}$
Next, we tackle the equation involving the two probabilities of interest starting with the probability on the left-hand side.
$\begin{align*}
s,t \ge 0, P(X\gt s+t\vert X\gt t) &= \frac{P(X\gt s+t, X\gt s)}{P(X\gt t)} \qquad \biggl[\ \text{since}\ P( A\vert B) = \frac{P(A,B)}{P(B)}\ \biggr] \\
&= \frac{P(X\gt s+t)}{P(X\gt t)} \\
& \qquad \biggl[\ \text{since}\ s,t\ge 0\implies s+t \ge s \implies\{X\gt s\}\cap\{X\gt s+t\} = \{X\gt s+t\}\ \biggr] \\
&= \frac{1-F_X(s+t)}{1-F_X(t)} \\
&= \frac{1-(1-e^{(s+t)/\lambda})}{1-(1-e^{-t/\lambda})} \\
&= \frac{e^{-s/\lambda-t/\lambda}}{e^{-t/\lambda}} \\
&= \frac{e^{-s/\lambda}e^{-t/\lambda}}{e^{-t/\lambda}} \\
&= e^{-s/\lambda} \\
&= 1-(1-e^{-s/\lambda}) \\
&= 1 - F_X(s) \\
s,t \ge 0, P(X\gt s+t\vert X\gt t) &= P(X\gt s).
\end{align*}$
Thus, $X$ indeed has the $\it{memoryless\ property}$.
