Short answer:
The expectation $\mathbb{E}_t$ in the PPO paper is not an expectation over trajectories, but a mean. I suppose the confusion comes from there. The two quantities are otherwise very similar. The gradient is however not the same, as it is a biased estimator.
Long answer:
There are many related formulations for RL problems, on one hand we have the finite-horizon (episodic) setting:
$$ J_1(\theta)=\mathbb{E}_{\tau\sim p_{\pi_{\theta}}}\left[\sum_{t= 0}^T r(s_t,a_t)\right]. $$
On the other hand, we have the commonly used infinite-horizon setting (assumed by the PPO paper):
$$ J_2(\theta)=\mathbb{E}_{\tau\sim p_{\pi_{\theta}}}\left[\sum_{t\geq 0} \gamma^tr(s_t,a_t)\right]. $$
In the finite horizon setting, we have :
\begin{equation} \nabla_{\theta} J_1(\theta) = \mathbb{E}_{\tau\sim p_{\pi_{\theta}}}\left[ \left(\sum_{t= 0}^T \nabla_{\theta}\log\pi_{\theta}(a_t\mid s_t) \right)R(\tau) \right], \end{equation}
where $R(\tau)=\sum_{t= 0}^T r(s_t,a_t)$. As explained in the SpinningUp post, we can replace $R(\tau)$ by $A_{\pi_{\theta}}(s_t,a_t)$.
Now, a similar proof shows that:
\begin{equation} \nabla_{\theta} J_2(\theta) = \mathbb{E}_{\tau\sim p_{\pi_{\theta}}}\left[ \sum_{t\geq 0}\gamma^t \nabla_{\theta}\log\pi_{\theta}(a_t\mid s_t) A_{\pi_{\theta}}(s_t,a_t) \right]. \end{equation}
Now, in practice it is very common to approximate
$$\nabla_{\theta} J_2(\theta) \approx \mathbb{E}_{\tau\sim p_{\pi_{\theta}}}\left[ \sum_{t\geq 0}\nabla_{\theta}\log\pi_{\theta}(a_t\mid s_t) A_{\pi_{\theta}}(s_t,a_t) \right]:=g^{\gamma},$$
that is we drop the discount factor $\gamma^t$. This relates to the policy gradient theorem and that we want to sample states from the steady-state distribution instead of the discounted state distribution.
Now, we want to approximate $A_{\pi}(s_t,a_t)$ by an estimator $\hat{A_t}$. Schulman et al in the GAE paper introduce the definition of a $\gamma$-just estimator. The exact definition is maybe not essential, but $\gamma$-just estimators include $Q_{\pi}(s_t,a_t)$, $A_{\pi}(s_t,a_t)$ and the GAE estimator used in the PPO paper.
So, if $\hat{A}_t$ is $\gamma$-just, then we have:
$$g^{\gamma} = \mathbb{E}_{\tau\sim p_{\pi_{\theta}}}\left[ \sum_{t\geq 0}\nabla_{\theta}\log\pi_{\theta}(a_t\mid s_t) \hat{A}_t \right]. $$
This expectation is however still not convenient for a practical algorithm, as we need to collect the trajectories with a policy with older weights $\pi_{\theta_{\text{old}}}$. So we approximate
$$g^{\gamma} \approx \mathbb{E}_{\tau\sim p_{\pi_{\theta_{\text{old}}}}}\left[ \sum_{t\geq 0}\nabla_{\theta}\log\pi_{\theta}(a_t\mid s_t) \hat{A}_t \right]:=\hat{g}. $$
Now, we need to approximate this expectation, so we collect $N$ rollouts of identical length $T$ (for simplicity) with the policy $\pi_{\theta_{\text{old}}}$, then we have an approximation of $g^{\gamma}$:
$$\nabla_{\theta} J_2(\theta) \approx g^{\gamma}\approx \hat{g}\approx\frac{1}{N}\sum_{n=0}^N\sum_{t=0}^T \nabla_{\theta}\log\pi_{\theta}(a_t^n\mid s_t^n) \hat{A}_t^n,$$
which is what the $\mathbb{E}_t$ in the PPO paper means. The expression on the right-hand side is now finally useful for gradient descent-like algorithms, as $\hat{A_t}$does not depend on $\theta$.
Note also that much of the PPO paper aims to ensure that the last approximation by replacing $p_{\pi_{\theta}}$ by $p_{\pi_{\theta_{\text{old}}}}$ is valid.
