No, your starting function is $y = Ke^{mx} + J$ and your last is $y = Ke^{mx} + K$. Unless $K = J$, these are clearly different functions.
When you divided both sides by $K$, you should get
$\frac{y}{K} = e^{mx} + \frac{J}{K}.$
Korgan, since you still don't understand, here are all the details. Let's start at the equation just above. If we want to replace $\frac{J}{K}$ with some other constant, let's use a different letter so it's less confusing. So, let $I = \frac{J}{K}$. This gives
$\frac{y}{K} = e^{mx} + I$
but we still have $K$ in the denominator, and never have we found any justification to know that $K = I$ so we can't just replace $K$ with $I$. After all, $I = \frac{J}{K}$. We never said anything about $I = K$. If you want to solve for $K$ to get rid of it, you have to solve $I = \frac{J}{K}$ for $K$ which gives $K = \frac{J}{I}$. So, our new equation reads
$\frac{y}{\frac{J}{I}} = e^{mx} + I$
Now, if you really want to call the new constant $K$, now would be a less confusing time to change it to $K$, so we would have
$\frac{y}{\frac{J}{K}} = e^{mx} + K$
Multiplying both sides by $\frac{J}{K}$ to clear the denominator gives
$y = \frac{J}{K} e^{mx} + J$
Notice, it STILL does not give
$y = K e^{mx} + K$
This would only happen in the specific case where $\frac{J}{K} = K$ and $J = K$, which would lead to $J = K = \pm 1$. But, we don't know $J$ or $K$ so we can't assume their values to make things simpler. They're unknown, yet fixed. They're not waiting around for us to simply pick their value.
And, also notice that where we end up is actually more complicated than where we started.