I start this answer with some consideration about the call option first:

When interest rates or volatility are not zero an American call option on a stock without dividend should never be exercised early. The proof is well known, easy and in fact similar to your proof for the put: the payoff the option holder gets from exercising the call option is \begin{align} S_t-E&\color{red}{\le}S_t-e^{-r(T-t)}E\le\underbrace{\max(S_t-e^{-r(T-t)}E,0)}_{\text{intrinsic option price}}\\ &\color{red}{\le}\text{European call price if you don't exercise}\\ &\le\text{American call price if you don't exercise}\,. \end{align} When $r>0$ or $\sigma>0$ at least one of the red $\color{red}\le$-signs is a strict inequality. This means never exercise early as you would gain $S_t-E$ in cash which is worth less than the remaining American option price.

To emphasize: even for the American call without dividends it may be optimal to exercise early, namely when $r=0$ and $\sigma=0$ so that $$\tag{1} S_t-E=\text{American call price if you don't exercise}. $$ In general ($r\ge 0,\sigma\ge 0$) this relationship is known as the rule exercise when $S_t$ reaches the exercise boundary.

What you have shown for the Put is

$$ E-S_t\color{red}{\le}\text{American put price if you don't exercise}\,. $$ To make the proof complete you would need to show that you never have an equals sign in this inequality.

I start this answer with some consideration about the call option first:

When interest rates or volatility are not zero an American call option on a stock without dividend should never be exercised early. The proof is well known, easy and in fact similar to your proof for the put: the payoff the option holder gets from exercising the call option is \begin{align} S_t-E&\color{red}{\le}S_t-e^{-r(T-t)}E\le\underbrace{\max(S_t-e^{-r(T-t)}E,0)}_{\text{intrinsic option price}}\\ &\color{red}{\le}\text{European call price if you don't exercise}\\ &\le\text{American call price if you don't exercise}\,. \end{align} When $r>0$ or $\sigma>0$ at least one of the red $\color{red}\le$-signs is a strict inequality. This means never exercise early as you would gain $S_t-E$ in cash which is worth less than the remaining American option price.

To emphasize: even for the American call without dividends it may be optimal to exercise early, namely when $r=0$ and $\sigma=0$ so that $$\tag{1} S_t-E=\text{American call price if you don't exercise}.^1 $$ In general ($r\ge 0,\sigma\ge 0$) this relationship is known as the rule exercise when $S_t$ reaches the exercise boundary.

What you have shown for the Put is

$$ E-S_t\color{red}{\le}\text{American put price if you don't exercise}\,. $$ To make the proof complete you would need to show that you never have an equals sign in this inequality.

$^1\quad $ When $r=\sigma=0$ "optimal" is to be taken with a grain of salt because (1) holds either for all $t$ or for no $t\,.$ If (1) holds with $r=\sigma=0$ one could therefore exercise whenever one wants.