Sorry I was not aware that you have the early stopping rule. As @Accumulation answered, your rule already use all $\alpha$ budget. If you decide whether you reject the null or not based on the second p-value only, your rejection rule is to reject the null if $(p_1, p_2) \in [0, \alpha_1] \times [0,1] \cup [0,1] \times [0, \alpha_2]$ where $\alpha_1$ is the first threshold you used to decide stop or continue and $\alpha_2$ is the threshold for the second p-value. In this case, it indues $\alpha_1 + \alpha_2 - \alpha_1 \alpha_2 = \alpha_1 + \alpha_2 (1- \alpha_1) \leq \alpha_1 + \alpha_2$ size test. This is the reson why I said all your $\alpha$ budget was used.
In this case, your combined p-value is given as
$$
p = p_1 I(p_1 \leq \alpha_1) + \left\{\alpha_1 + p_2(1-\alpha_1)\right\}I(p_1 > \alpha_1).
$$
Therefore, once $p_1 > \alpha_1$, your p-value is greater than or equal to $\alpha_1$ regardless of $p_2$ value.
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[Previous answer - I was not aware the stopping rule]
I assume you are working with a non-conservative null, that is, under the null, the p-value has the uniform distribution on $[0,1]$. Since your p-values are two independent samples from $U[0,1]$ under the null, you can use any "data-independent" rejection region on $[0,1]^2$ whose area is less than your test level $\alpha$. (Most natural one would be $[0, \sqrt{\alpha}]^2$ and, in this case, p-value is $(\max\{p_1, p_2\})^2$)
However, if you can calculate a new p-value based on $S_1 \cup S_2$, you can also use this p-value as a merged p-value since you stopped trials in a data-independent manner (out-of-time).