I believe it should be the same. The basic formula for the independent two-sample T-test is: $$ t = \frac{\bar{X}_{1} - \bar{X}_{2}}{s_{p}\times\sqrt{\frac{2}{n}}} $$

If you did the z-score standardization, but have not changed the data otherwise. It is obvious that $\sqrt{\frac{2}{n}}$ is unchanged. So we just need to make sure that the ratio between the numerator and the denominator is unchanged too. Let's start with the denominator. The pooled standard deviation $s_p$ is:

$$ s_{p} = \sqrt{\frac{1}{2}\times(\sigma_{x_1}^{2}+\sigma_{x_2}^{2})} $$

Where $\sigma$ is the variance of the group:

$$ \sigma_{x_{1}}^{2} = \frac{\sum_{i=1}^{n}{(x_{i}-\bar{X_{1}})^{2}}}{{n-1}} $$

It can be assumed again that $n$ haven't changed. How much the sum part have changed due to standardization? For that let's look at the z-score formula: $$ z = x-\bar{x} \times \frac{n-1}{\sum{x-\bar{x}}} $$ That's a transformation that we apply to every element in our initial dataset. The critical parts are $x_i - \bar{x}$ from here and $\bar{X_1} - \bar{X_2}$ from the t-stat formula, as $n$ is unchanged. What we need to make sure essentially - to prove that the t statistics is the same - that these expressions have the same ratio in the initial and the z-score case. This can be proven by showing that the ratio of the mean and the individual values (the relative distance from the mean) is unchanged after the z-score transformation. Essentially:

$$ \frac{x_{i}}{\bar{x}} = \frac{z_{i}}{\bar{z}} $$

and this equation holds (see proof below) - the z-score doesn't change the relative distance between the values and the mean, actually it shows the distance from the mean in $\sigma$ units. Even if the actual values of the $\sigma$s will change their relative position to each other won't. That's kind of the point of the standardization - keep the distances, but lose the original level.

So back to the original t-statistic: $$ t = \frac{\bar{X}_{1} - \bar{X}_{2}}{s_{p}\times\sqrt{\frac{2}{n}}} $$ As individual values keep a relative distance from the mean, $\bar{X_1} - \bar{X_2}$ will be different from $\bar{Z_1} - \bar{Z_2}$, but as we've changed the pooled standard deviation (because of $x_i - \bar{X_1}$) with the same scale once we move into calculating relative measures we end up with the same results.

Proof: $$ \frac{x_{i}}{\frac{\sum{x_{i}}}{n}} = \frac{ \frac{x_{i}-\bar{x}}{\sigma} }{\frac{\sum{\frac{x_{i}-\bar{x}}{\sigma}}}{n}} $$ $$ \sum{\frac{x_i-\bar{x}}{\sigma}} \times \frac{x_i}{\sum{x_i}} = \frac{x_i-\bar{x}}{\sigma} $$ $$ \frac{x_{i}}{\sigma}\times\sum{x_i-\frac{x_i}{\sum{x_i}}}\times\frac{1}{\sum{x_i}} = \frac{x_{i}}{\sigma}(1-\frac{1}{\sum{x_i}}) $$ $$ \frac{\sum{x_i-\frac{x_i}{\sum{x_i}}}}{x_i} = 1 - \frac{1}{\sum{x_i}} $$ where simplifying the LHS leaves us with $$ 1 - \frac{1}{\sum{x_i}} = 1 - \frac{1}{\sum{x_i}} $$ Thus proving that: $$ \frac{x_{i}}{\bar{x}} = \frac{z_{i}}{\bar{z}} $$

I believe it should be the same.

Short answer: z-score standardization is a linear transformation and as such won't change the ratio that's the basis of the T-test.

Long: The basic formula for the independent two-sample T-test is: $$ t = \frac{\bar{X}_{1} - \bar{X}_{2}}{s_{p}\times\sqrt{\frac{2}{n}}} $$

If you did the z-score standardization, but have not changed the data otherwise. It is obvious that $\sqrt{\frac{2}{n}}$ is unchanged. So we just need to make sure that the ratio between the numerator and the denominator is unchanged too. Let's start with the denominator. The pooled standard deviation $s_p$ is:

$$ s_{p} = \sqrt{\frac{1}{2}\times(\sigma_{x_1}^{2}+\sigma_{x_2}^{2})} $$

Where $\sigma$ is the variance of the group:

$$ \sigma_{x_{1}}^{2} = \frac{\sum_{i=1}^{n}{(x_{i}-\bar{X_{1}})^{2}}}{{n-1}} $$

It can be assumed again that $n$ haven't changed. How much the sum part have changed due to standardization? For that let's look at the z-score formula: $$ z = x-\bar{x} \times \frac{n-1}{\sum{x-\bar{x}}} $$ That's a transformation that we apply to every element in our initial dataset. The critical parts are $x_i - \bar{x}$ from here and $\bar{X_1} - \bar{X_2}$ from the t-stat formula, as $n$ is unchanged. What we need to make sure essentially - to prove that the t statistics is the same - that these expressions have the same ratio in the initial and the z-score case. This can be proven by showing that the ratio of the mean and the individual values (the relative distance from the mean) is unchanged after the z-score transformation. Essentially:

$$ \frac{x_{i}}{\bar{x}} = \frac{z_{i}}{\bar{z}} $$

and this equation holds (see proof below) - the z-score doesn't change the relative distance between the values and the mean, actually it shows the distance from the mean in $\sigma$ units. Even if the actual values of the $\sigma$s will change their relative position to each other won't. That's kind of the point of the standardization - keep the distances, but lose the original level.

So back to the original t-statistic: $$ t = \frac{\bar{X}_{1} - \bar{X}_{2}}{s_{p}\times\sqrt{\frac{2}{n}}} $$ As individual values keep a relative distance from the mean, $\bar{X_1} - \bar{X_2}$ will be different from $\bar{Z_1} - \bar{Z_2}$, but as we've changed the pooled standard deviation (because of $x_i - \bar{X_1}$) with the same scale once we move into calculating relative measures we end up with the same results.

Proof: $$ \frac{x_{i}}{\frac{\sum{x_{i}}}{n}} = \frac{ \frac{x_{i}-\bar{x}}{\sigma} }{\frac{\sum{\frac{x_{i}-\bar{x}}{\sigma}}}{n}} $$ $$ \sum{\frac{x_i-\bar{x}}{\sigma}} \times \frac{x_i}{\sum{x_i}} = \frac{x_i-\bar{x}}{\sigma} $$ $$ \frac{x_{i}}{\sigma}\times\sum{x_i-\frac{x_i}{\sum{x_i}}}\times\frac{1}{\sum{x_i}} = \frac{x_{i}}{\sigma}(1-\frac{1}{\sum{x_i}}) $$ $$ \frac{\sum{x_i-\frac{x_i}{\sum{x_i}}}}{x_i} = 1 - \frac{1}{\sum{x_i}} $$ where simplifying the LHS leaves us with $$ 1 - \frac{1}{\sum{x_i}} = 1 - \frac{1}{\sum{x_i}} $$ Thus proving that: $$ \frac{x_{i}}{\bar{x}} = \frac{z_{i}}{\bar{z}} $$