Everything that you have written is correct. You can always test out things like that with a toy example. Here is an example with R:
library(MASS)
rho <- .5 ### the true correlation in both groups
S1 <- matrix(c( 1, rho, rho, 1), nrow=2)
S2 <- matrix(c(16, 4*rho, 4*rho, 1), nrow=2)
cov2cor(S1)
cov2cor(S2)
xy1 <- mvrnorm(1000, mu=c(0,0), Sigma=S1)
xy2 <- mvrnorm(1000, mu=c(0,0), Sigma=S2)
x <- c(xy1[,1], xy2[,1])
y <- c(xy1[,2], xy2[,2])
group <- c(rep(0, 1000), rep(1, 1000))
summary(lm(y ~ x + group + x:group))
What you will find that the interaction is highly significant, even though the true correlation is the same in both groups. Why does that happen? Because the raw regression coefficients in the two groups reflect not only the strength of the correlation, but also the scaling of X (and Y) in the two groups. Since those scalings differ, the interaction is significant. This is an important point, since it is often believed that to test the difference in the correlation, you just need to test the interaction in the model above. Let's continue:
summary(lm(xy2[,2] ~ xy2[,1]))$coef[2] - summary(lm(xy1[,2] ~ xy1[,1]))$coef[2]
This will show you that the difference in the regression coefficients for the model fitted separately in the two groups will give you exactly the same value as the interaction term.
What we are really interested in though is the difference in the correlations:
cor(xy1)[1,2]
cor(xy2)[1,2]
cor(xy2)[1,2] - cor(xy1)[1,2]
You will find that this difference is essentially zero. Let's standardize X and Y within the two groups and refit the full model:
x <- c(scale(xy1[,1]), scale(xy2[,1]))
y <- c(scale(xy1[,2]), scale(xy2[,2]))
summary(lm(y ~ x + x:group - 1))
Note that I am not including the intercept or the group main effect here, because they are zero by definition. You will find that the coefficient for x is equal to the correlation for group 1 and the coefficient for the interaction is equal to the difference in the correlations for the two groups.
Now, for your question whether it would be better to use this approach versus using the test that makes use of Fisher's r-to-z transformation.
EDIT
The standard errors of the regression coefficients that are calculated when you standardize the X and Y values within the groups do not take this standardization into consideration. Therefore, they are not correct. Accordingly, the t-test for the interaction does not control the Type I error rate adequately. I conducted a simulation study to examine this. When $\rho_1 = \rho_2 = 0$, then the Type I error is controlled. However, when $\rho_1 = \rho_2 \ne 0$, then the Type I error of the t-test tends to be overly conservative (i.e., it does not reject often enough for a given $\alpha$ value). On the other hand, the test that makes use of Fisher's r-to-z transformation does perform adequately, regardless of the size of the true correlations in both groups (except when the group sizes get very small and the true correlations in the two groups get very close to $\pm1$.
Conclusion: If you want to test for a difference in correlations, use Fisher's r-to-z transformation and test the difference between those values.
Best Answer
The equation you are using is the Michaelis-Menten equation. In 1934 Lineweaver and Burk wrote a linear transform for this equation by taking the reciprocal of both sides or
1/v = 1/Vmax +Km/V*S
There are other linear transforms for the Michaelis Menten equation, See "Enzyme Kinetics" by Segel. Now it is usually preferable to do a linear transform when doing an analysis of of this sort. The R code is easier and you can simply use an analysis of variance to see if the slopes of your lines are equal. Also, when you do a linear transform, plot the lines. If they do not appear straight, then the system you are modeling is not strictly Michaelis-Menten. You have the wrong model.