For an American option, we are faced with the problem of determining the optimal exercise level/boundary $S_*$. For a put, we clearly want $S_t < K$ in order to have a positive payoff. The question is how much less than $K$? Zero could be a choice but that might be a long time to wait. Considering the time value of money that could be sub optimal. Hence for a put we define $S_* < K$ as our optimum exercise level.
Conversely for a call, we want $S$ to be much larger than $K$. So what would be an optimum exercise level?

Perpetual American Option with Dividends. $T=\infty$

In the Black & Scholes setting, $\partial_t V = 0$ because it does not depend on time anymore! Hence
$$\dfrac{1}{2}\sigma^2 S^2 \dfrac{d^2V}{dS^2}(S) + (r-q) S \dfrac{dV}{dS}(S) = rV (S) \tag{1}$$

Remember! An American option on a non-dividend paying asset ($q = 0$) is the same as a European option. And since for a perpetual option the option is never exercised, it is the same as holding the stock.

Observe that the partials disappear as we have one variable since $\dfrac{\partial V}{\partial t} = 0$. (Hence the single argument $S$ is shown for emphasis). The equation must be satisfied for

$S > S^*$ for a call. $V(S,S^*) \equiv C^A(S^*) = \text{max} (S^* – K), 0 = S^* – K\tag{2a}$
$S < S_*$ for a put. $V(S,S^*) \equiv P^A(S^*) = \text{max} (K – S^* ), 0 = K – S^* \tag{2b}$

Additionally as we seek to maximize the value of the option for the holder,

$\dfrac{dC^A}{dS}\bigg{|}_{(S=S^*)} = 1\tag{3a}$ for a call
$\dfrac{dP^A}{dS}\bigg{|}_{(S=S^*)} = -1\tag{3b}$ for a put

Collectively, $(1), (2)$ and $(3)$ constitute the free boundary problem for American calls and puts within the construct of infinite maturity $T$. $(3)$ is known as the smooth pasting or tangency condition. $(1)$ is an Euler type linear equidimensional[ref]Mathematical Models of Financial Derivatives. Y. K. Kwok. Chapter 5.1.3. Page 259[/ref] ordinary differential equation. For such an ODE, the general solution is of the form $$V(S;S^*) = C_{+}S^{\lambda +} + C_{-}S^{\lambda -}\tag{4}$$ where $\lambda_{\pm}$ are the roots of the auxilliary equation obtained by substituting $S^{\lambda}$ back into $(1)$.
$$dS^\lambda = \lambda S^{\lambda – 1}\; \Rightarrow \; S \dfrac{dS^\lambda}{dS} = \lambda S^\lambda$$ $$\dfrac{d^2S^\lambda}{dS^2} = \lambda(\lambda – 1) S^{\lambda – 2}\; \Rightarrow \; S^2 \dfrac{d^2S^\lambda}{dS^2} = \lambda(\lambda – 1)S^\lambda$$ Hence substituting into $(1)$ $$\dfrac{1}{2}\sigma^2 \lambda(\lambda – 1){S^\lambda} + (r-q)\lambda{S^\lambda} \;-\; r{S^\lambda} = 0$$ $$\dfrac{1}{2}\sigma^2 \lambda(\lambda – 1) + (r-q)\lambda \;-\; r = 0$$ which factorises to the auxilliary quadratic equation in $\lambda$ $$\dfrac{\sigma^2}{2} \lambda^2 + (r-q -\dfrac{\sigma^2}{2} )\lambda \;-\; r = 0$$ $$\lambda_{\pm} = \dfrac{-(r-q -\dfrac{\sigma^2}{2} ) \pm \sqrt{\biggl(r-q -\dfrac{\sigma^2}{2} \biggr)^2 + 2 \sigma^2 r}}{\sigma^2}$$ since $\sigma > 0$, and $r > 0$ and since $(r-q -\dfrac{\sigma^2}{2} + 2 \sigma^2 r) > (r-q -\dfrac{\sigma^2}{2}) $ we can observe that $$\lambda_- <0$$ $$\lambda_+ >0$$The signs of the roots are important!

Determine constants $C_{+}, C_{-}$ and $S_{*}$

Now recall $(4)$ $$V(S;S^*) = C_{+}S^{\lambda +} + C_{-}S^{\lambda -}$$

we propose that coefficients $C_{+}$ and $C_{-}$ are not zero

Call Option: We want $V(S < S^*) \rightarrow 0$ as $S \rightarrow 0$. Clearly with the negative root $S^{\lambda_-}$ this will not happen. So the term $C_{-}S^{\lambda_-}$ must be zero for a call. Hence \begin{align*}V_{\text{call}}(S < S^*)&= C_{+}S^{\lambda_+}\\ &=S_* - K\\ \dfrac{d}{dS}V_{\text{call}} \bigg{|}_{S*} &= \lambda_+ C_{+}S_{*}^{\lambda_+ -1} = 1 \end{align*} i.e the derivative of the call payoff $= 1$

Put Option: We want $V(S > S_{*}) \rightarrow 0$ as $S \rightarrow \infty$. Clearly with the positive root $S^{\lambda_+}$ this will not happen. So the term $C_{+}S^{\lambda_+}$ must be zero for a put. Hence

\begin{align*}V_{\text{put}}(S > S_{*}) &= C_{-}S^{\lambda_-}\\
&=K – S_*\\
\dfrac{d}{dS}V_{\text{put}} \bigg{|}_{S*} &= \lambda_- C_{-}S_{*}^{\lambda_- -1} = -1
\end{align*}
i.e the derivative of the put payoff $= -1$

Note: If alternatively, we may ask, when does $V(S;S^*)$ approach infinity OR some bounded value given the polarity of the roots?$$\lambda_- (<0): \quad V_{\text{put}}(S;S^*) \rightarrow \text{Max (Bounded) if } \; S^{\lambda_-} \rightarrow 0 $$
And not if $S^{\lambda_+} \rightarrow 0 $. Hence, the $C_{+} S^{\lambda_+}$ term must be zero for $V_{\text{put}}(S;S^*)$

for puts, most texts look at what happens as $S \rightarrow \infty$

For a call, the direct relationship is obvious:
$$\lambda_+ (>0):\quad V_{\text{call}}(S < S^*) \rightarrow \infty \; \text{ if } \; S^{\lambda_+} \rightarrow \infty$$ So the term $C_{-}S^{\lambda_-}$ must be zero for a call.