tldr If $\mathbb{E}[\varepsilon_{i,t}|X_{i,t-1}, Y_{i,t-1}] = 0$ then the coefficient $\beta_2$ is equal to:
$$
\frac{\partial \mathbb{E}[\ln Y_{i,t}|X_{i,t-1}= x_{i,t-1}, Y_{i,t-1} = y_{i,t-1}]}{\partial \ln x_{i,t-1}}
$$
If $\varepsilon_{i,t}$ is independent of $X_{i,t-1}$ and $Y_{i,t-1}$ (which is a stronger condition) then it is also equal to:
$$
\frac{\partial \ln \mathbb{E}[Y_{i,t}|X_{i,t-1}= x_{i,t-1}, Y_{i,t-1} = y_{i,t-1}]}{\partial \ln x_{i,t-1}}
$$
In both cases, they give an estimate for the elasticity of employment with respect to productivity yesterday (conditional on employment yesterday), i.e. what is the (expected) percentage point change in employment (today) due to a 1 percentage increase in productivity (yesterday) given the level of employment yesterday.
estimating elasticities
If
$$
\ln y = \alpha + \beta \ln x,
$$
Then the $x$-elasticity of $y$ is given by:
$$
\frac{\partial \ln y}{\partial \ln x} = \beta.
$$
It measures the percentage change in $y$ due to a 1 percent increase in $x$.
Now let's go to the stochastic framework:
$$
\ln Y = \alpha + \beta \ln X + \varepsilon
$$
And assume, as usual, that $\mathbb{E}[\varepsilon] = 0$.
The elasticity of $Y$ with respect ot $X$ is not really defined as $Y$ is not a function of $X$ (i.e. both are random variables).
There are two natural generalizations of elasticity in this setting:
- $\dfrac{\partial \ln \mathbb{E}[Y|X = x]}{\partial \ln x}$, which is the elasticity of the conditional mean function $\mathbb{E}[Y|X = x]$.
- $\dfrac{\partial \mathbb{E}[\ln Y|X = x]}{\partial \ln x}$, which is the derivative of the conditional mean function $\mathbb{E}[\ln Y|X = x]$.
Because the log is a non-linear transformation, the two are not necessarily the same.
However, if $X$ and $\varepsilon$ are independent then they are identical and equal to $\beta$. To see this, notice that for 1.
$$
Y = e^\alpha X^\beta e^\varepsilon
$$
Then taking conditional expecations and using the independence between $\varepsilon$ and $X$:
$$
\mathbb{E}[Y|X = x] = e^\alpha x^\beta \mathbb{E}[e^\varepsilon|X = x] = e^\alpha X^\beta \mathbb{E}[e^\varepsilon],\\
\to \ln \mathbb{E}[Y|X = x] = \alpha + \beta \ln x + \ln \mathbb{E}[e^\varepsilon].
$$
Taking the partial derivative with respect to $\ln(x)$ gives:
$$
\frac{\partial \ln \mathbb{E}[Y|X = x]}{\partial \ln x} = \beta.
$$
For 2. we immediately have:
$$
\mathbb{E}[\ln Y|X = x] = \alpha + \beta \ln x + \mathbb{E}[\varepsilon|X = x] = \alpha + \beta \ln x.
$$
So:
$$
\frac{\partial E[\ln Y|X = x]}{\partial \ln x} = \beta.
$$
If only $\mathbb{E}[\varepsilon|X] = 0$ but $X$ and $\varepsilon$ are not independent, then only the derivation of 2 is valid and 1 and 2 are not necessarily equal.
If you want to estimate the elasticity of the growth rate, then the right regression is indeed:
$$
\ln(Y_{i,t}/Y_{i,t-1}) = \beta_0 + \beta_1 \ln X_{i,t} + \varepsilon_{i,t}.
$$