Let $X \sim N(\mu,\sigma^2)$ and $\{x_i\}$ and $\{y_i\}$ be two same size i.i.d samples from $X$.
Now we do z-tests (assuming variance is known) individually for the two samples and then together. In both cases the null hypothesis is same:
$$H_0: \mu=0$$
Let $p_x$ and $p_y$ be respective p-values for individual tests and $p_{xy}$ be the p-value for combined test:
We know that under null hpothesis $$\bar{X}, \bar{Y} \sim N(0,\sigma^2/n)$$
Now,
$$p_x=Pr\bigg(-\bigg|\frac{\bar{x}}{\sigma/\sqrt{n}}\bigg| \geq Z \geq \bigg|\frac{\bar{x}}{\sigma/\sqrt{n}}\bigg|\bigg) = 2\Phi\bigg(-\bigg|\frac{\bar{x}}{\sigma/\sqrt{n}}\bigg|\bigg)$$ where $\Phi(.)$ is the cdf for $N(0,1)$
So, $$p_xp_y=4\Phi\bigg(-\bigg|\frac{\bar{x}}{\sigma/\sqrt{n}}\bigg|\bigg)\Phi\bigg(-\bigg|\frac{\bar{y}}{\sigma/\sqrt{n}}\bigg|\bigg)$$
whereas in the combined test:
$$p_{xy}=2\Phi\bigg(-\bigg|\frac{(\bar{x}+\bar{y})/2}{\sigma/\sqrt{2n}}\bigg|\bigg)$$
Now if I understand your question, you want to know what if we reject null based on $p_xp_y$ as compared to when we reject based on $p_{xy}$.
Under a true null hypothesis, p-value is uniformly distributed on $[0,1]$. So,
$$Pr(p_{xy} \leq 0.05) = 0.05$$
However, same is not true for $p_xp_y$ as it is not uniformly distributed. It's cdf is $z-z\ln{z}$. See this for derivation.
$$Pr(p_xp_y \leq 0.05) \approx 0.2$$
So you are clearly rejecting the null hypothesis much more often leading to higher type I error.