= Solution
The <stationary points> of the exponent are $t=1/3$ and $t=1$, because $\phi'(t)=(3t-1)(t-1)$. The first is a local maximum with $\phi(1/3)=4/27$ and $\phi''(1/3)=-2$; the second is a local minimum. Also
$$
\phi(t)-\frac4{27}=(t-\tfrac13)^2(t-\tfrac43),
$$
so the endpoint overtakes the interior maximum only when $b>4/3$.
For fixed $0<b<1/3$, the upper endpoint is the unique maximum. Linearizing the exponent there gives the <Laplace endpoint estimate> $e^{\lambda\phi(b)}/(\lambda\phi'(b))$. At $b=1/3$, only half of the local <Gaussian integral> is included. For $1/3<b\leq4/3$, the whole Gaussian near $1/3$ contributes. At $b=4/3$ an endpoint has the same exponential height, but its prefactor is $\lambda^{-1}$, smaller than the interior $\lambda^{-1/2}$ prefactor. For $b>4/3$, the endpoint has strictly greater height and dominates.
Consequently \b[all fixed-$b$ regimes are]
$$
\boxed{I(\lambda)\sim
\begin{cases}
\dfrac{e^{\lambda b(b-1)^2}}{\lambda(3b^2-4b+1)},&0<b<\tfrac13,\\[6pt]
\dfrac{\sqrt\pi}{2\sqrt\lambda}e^{4\lambda/27},&b=\tfrac13,\\[6pt]
\dfrac{\sqrt\pi}{\sqrt\lambda}e^{4\lambda/27},&\tfrac13<b\leq\tfrac43,\\[6pt]
\dfrac{e^{\lambda b(b-1)^2}}{\lambda(3b^2-4b+1)},&b>\tfrac43.
\end{cases}}
$$
The estimate follows from <Laplace's method>; the tail toward negative infinity is exponentially harmless since the cubic tends to negative infinity.
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