= Solution
Let the <Poisson process> $X$ have rate $\lambda$, let $\overline F(t)=\mathbb P(\xi_1>t)$ be the survival function of an inter-renewal time of $N$, and let $\overline H$ be the survival function of the first event time of $Y$. Independence gives
$$
\boxed{\overline H(t)=e^{-\lambda t}\overline F(t)}.
$$
Put
$$
\mu_F=\int_0^\infty\overline F(t)\,dt,
\qquad
\mu_H=\int_0^\infty\overline H(t)\,dt.
$$
At a large time, the excess of $Y$ is the minimum of the excesses of $X$ and $N$. The Poisson excess is an independent rate-$\lambda$ exponential random variable. Applying the renewal excess limit theorem to $N$ and to the assumed renewal process $Y$ gives
$$
\boxed{
\frac1{\mu_H}\int_x^\infty\overline H(y)\,dy
=\frac{e^{-\lambda x}}{\mu_F}
\int_x^\infty\overline F(y)\,dy
}.
$$
Substituting $\overline F(y)=e^{\lambda y}\overline H(y)$ yields the integral equation
$$
\frac1{\mu_H}\int_x^\infty\overline H(y)\,dy
=\frac{e^{-\lambda x}}{\mu_F}
\int_x^\infty e^{\lambda y}\overline H(y)\,dy.
$$
Set $A(x)=\int_x^\infty\overline H(y)dy$ and $c=\mu_H/\mu_F<1$. Differentiating the equation almost everywhere gives
$$
(1-c)\overline H(x)=\lambda A(x).
$$
Since $A'=-\overline H$, it follows that
$$
A'(x)=-\frac{\lambda}{1-c}A(x).
$$
Using $\overline H(0)=1$ shows that $\overline H(x)=e^{-x/\mu_H}$. Thus the first event time of $Y$ is exponential. This is the <exponential first interarrival in a renewal superposition containing a Poisson process> theorem.
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