= Solution
For $i\geq1$, the total holding rate is $(\lambda+1)2^i$, and the <jump chain> moves upward and downward with probabilities
$$
p=\frac{\lambda}{\lambda+1},
\qquad
q=\frac1{\lambda+1},
$$
while it moves from zero to one with probability one. Thus it is a <reflected nearest-neighbour random walk>. Recurrence is unaffected by the holding times, so it is recurrent exactly when $p\leq q$. In the given range,
$$
\boxed{X\text{ is recurrent exactly when }\lambda=1.}
$$
For $\lambda>1$ the jump chain is transient.
The <detailed balance for a birth-death process>[detailed-balance equations] are
$$
\pi_i\lambda2^i=\pi_{i+1}2^{i+1}
\qquad(i\geq0),
$$
where the formula also covers the boundary $i=0$. Hence every candidate invariant measure has
$$
\pi_i=\pi_0\left(\frac\lambda2\right)^i.
$$
At $\lambda=1$ this is summable and the chain is nonexplosive, giving
$$
\pi_i=2^{-i-1},
\qquad i\geq0.
$$
For $\lambda=2$ it is not summable. When $1<\lambda<2$, it is a summable formal solution of $\pi Q=0$, but the chain explodes, so it is not an invariant probability for the minimal process; this is precisely the caveat in <invariant distribution of an explosive chain>. Consequently
$$
\boxed{X\text{ has an invariant distribution exactly when }\lambda=1.}
$$
It remains to classify explosion. At $\lambda=1$, the recurrent jump chain visits zero infinitely often. The holding time at zero has rate one, so <nonexplosion from recurrent visits to a slow state> shows that the accumulated time is infinite almost surely.
For $\lambda>1$, let $V_i$ be the total number of visits of the transient jump chain to $i$. Using the stated uniform visit bound and writing $q_i=(\lambda+1)2^i$ for $i\geq1$,
$$
\mathbb E\!\left[\sum_{n\geq0}\frac1{q_{Y_n}}\right]
=\sum_{i\geq0}\frac{\mathbb E V_i}{q_i}
<\infty,
$$
because $\sum_i2^{-i}<\infty$. Conditional on the jump path, this is the expected sum of all holding times, so the total lifetime is finite almost surely. This is <explosion of an upward-biased walk with geometrically increasing rates>. Therefore
$$
\boxed{X\text{ is explosive exactly when }1<\lambda\leq2.}
$$
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