Not quite.
Your interpretation for the first model is correct, but your explanation isn't quite right. Note that
$$ \begin{equation*} \beta_1 = \frac{\partial \log y}{\partial x}, \end{equation*}$$
but that isn't very easy to interpret. So, we recall the calculus result that
$$ \begin{equation*} \frac{\partial \log y}{\partial y} = \frac{1}{y} \end{equation*}$$
or
$$ \begin{equation*} \partial \log y = \frac{\partial y}{y}. \end{equation*}$$
Plugging this into the equation for $\beta_1$, we have
$$ \begin{equation*} \beta_1 = \frac{\partial y / y}{\partial x}. \end{equation*}$$
If we multiply both sides by 100, we have
$$ \begin{equation*} 100\beta_1 = \frac{100 \times \partial y / y}{\partial x}. \end{equation*}$$
We realize that $100 \times \partial y/y$ is just the percentage change in $y$, giving the interpretation that $100 \beta_1$ is the percentage change in the outcome for a one unit increase in $x$.
The correct interpretation for your second model would be that a 1 unit increase in GDP leads to a 10 percentage point increase in sales. It's easiest to understand this by thinking of your outcome as being measured in percentage points, rather than percent. Then, a 1 unit change in $x$ leads to a $\beta_1$ unit change in $y$, just as we normally get.
This is an important distinction. An increase in sales from 1% to 5% is a $5 - 1 = 4$ percentage point increase, but a $(5 - 1)/1 \times 100 =400$ percent change.
The question concerns models of the form
$$\log(y) = \cdots + \beta \log(x) + \cdots$$
(where none of the omitted terms involves anything that changes with $x$). When we change $x$ by $100\delta\%$ we multiply it by $1+\delta$. According to the laws of logarithms, when $1 + \delta \gt 0$,
$$\log(x(1+\delta)) = \log(x) + \log(1 + \delta).$$
Therefore if such a change in $x$ changes $y$ to $y^\prime$,
$$\eqalign{
\log(y^\prime) &= \cdots + \beta \log(x(1+\delta)) + \cdots \\
&= \cdots + \beta\left(\log(x) + \log(1 + \delta)\right) + \cdots \\
&= \cdots + \beta\log(x) + \beta\log(1 + \delta) + \cdots .
}$$
The change in $\log(y)$ is $$\log(y^\prime) - \log(y) = \beta\log(1 + \delta).$$
When $\delta$ is nearly zero (say, $10\%$ or smaller in size), $\log(1 + \delta) \approx \delta$ is a good approximation. When in turn $\beta\delta$ is also close to zero, this is the basis for the approximate interpretation "a $\delta$ percent change in $x$ corresponds to a $\beta\delta$ percent change in $y$."
For larger $\delta$ or $\beta$, however, this approximation fails. The fully general relationship is obtained by noting that adding $\beta\log(1+\delta)$ to $\log{y}$ is tantamount to multiplying $y$ by
$$\exp(\beta\log(1+\delta)) = (1+\delta)^\beta.$$
Therefore, when working with logarithms, think multiplicatively. We may memorialize the result of this analysis with a simple rule:
When $x$ is multiplied by any positive amount $c$, $y$ is multiplied by $c^\beta$.
In other words, log-log relationship are power relationships. Let's look at some examples:
When $\beta=2$, multiplying $x$ by $c$ multiplies $y$ by $c^2$. For instance, tripling $x$ will multiply $y$ by $9$.
When $\beta=1/3$, multiplying $x$ by $c$ multiplies $y$ by $c^{1/3} = \sqrt[3]{c}$. For instance, doubling $x$ will only multiply $y$ by $\sqrt[3]{2}\approx 1.26$.
When $\beta = -1$, multiplying $x$ by $c$ multiplies $y$ by $c^{-1} = 1/c$; that is, $y$ is divided by $c$.
Best Answer
If $ln(p_t)=\beta_0 + \beta_1 (ln(E_t)-ln(E_{t-1}))$ then it holds that $ln(p_t)=\beta_0 + \beta_1 ln\left(\frac{E_t}{E_{t-1}}\right)$ or $ln(p_t)=\beta_0 + ln\left( \left(\frac{E_t}{E_{t-1}}\right)^{\beta_1}\right)$.
After 'exponentiation' this becomes $p_t=e^{\beta_0}\left(\frac{E_t}{E_{t-1}}\right)^{\beta_1}$.
So if $\frac{E_t}{E_{t-1}}$ is multiplied by a factor $f$, i.e. $\frac{E_t}{E_{t-1}} \to f \frac{E_t}{E_{t-1}}$, then the new $p_t$ will become $e^{\beta_0}\left(f\frac{E_t}{E_{t-1}}\right)^{\beta_1}=f^{\beta_1}e^{\beta_0}\left(\frac{E_t}{E_{t-1}}\right)^{\beta_1}=f^{\beta_1}p_t$.
So if $\frac{E_t}{E_{t-1}} \to f \frac{E_t}{E_{t-1}}$ then the price $p_t$ will become $p_t \to f^{\beta_1}p_t$.
For example, if $\frac{E_t}{E_{t-1}}$ increases by $1\%$ then $f=1.01$ and the price then increases by $(1.01^{\beta_1} - 1) \times 100 \%$.
Note that, for very small $x$ it holds that ( Taylor series expansion) $(1+x)^{\beta_1} \approx 1+ \beta_1 x$.
Applying this approximation to the above example we find that $(1.01)^{\beta_1}=(1+0.01)^{\beta_1} \approx 1+ \beta_1 \times 0.01$ so the increase in $p$ is $(1.01^{\beta_1} - 1) \times 100 \% \approx (1+\beta_1 \times 0.01 -1) \times 100\%=100 \times 0.01 \times \beta_1 \%=\beta_1 \%$.
So your interpretation is fine with the nuance that you talk about relative changes in growth rate, so if the growth rate is $\frac{E_t}{E_{t-1}}= 5\%$ then a 10% change brings this growth rate to 5.5% ($=1.1 \times 5\%$) , a 1% change brings it to 5.05% ($=1.01 \times 5\%$) (moreover, the approximation only holds for small percentages)