These regressions model the expected value of a random variable as some function of the dependent variables. So if you take the $\log$ of $Y$ and assume a main effects model then the model is:
$$ E(\log Y) = \beta_0 + \beta_1 x_1 + ... + \beta_p x_p $$
Since the $E()$ and $\log$ cannot be interchanged, this is not the same as
$$ E(Y) = \exp(\beta_0 + \beta_1 x_1 + ... + \beta_p x_p) $$.
So saying 'transforming the dependent variable with a log may be equivalent to transforming the sum of independent variables with an exponential link function' is not correct - for two reasons 1) E() and log cannot be interchanged and 2) you don't just have the sum of independent variables, you have a linear combination of them.
However, both of the models above, can be thought of as GLM's where you model the mean as a linear function of the dependent variables ($\mu = g^{-1}(X\beta)$). The $\mu$ in the first represents the mean of $\log Y$ and in the second is the mean of $Y$. To complete the formulation, the distribution of the respective variable is specified. As stated in the comments, assuming $\log Y$ is normal is equivalent to $Y$ being log-normal.
As far as GAMs are concerned - they model the mean as:
$$g(E(Z)) = \beta_0 + f_1(x_1) + f_2(x_2) + ... + f_m(x_m)$$
where Z is some random variable. So taking $g$ as the identity, the first could be considered a GAM on $Z = \log Y$. Even if some of the x's on the right side are actually the logs of the original dependent variables. For example, if you had $m = 2$ and :
$$ E(\log Y) = \beta_0 + \beta_1 \log x_1 + \beta_2 x_2 $$
Then $f_1(x_1) = \beta_1 \log x_1$ and $f_2(x_2) = \beta_2 x_2$. The second case wouldn't fall into the GAM framework because there is no way to write the right hand side as a sum of $p$ functions, each a function of only one dependent variable.
f_1()...f_n()
. $\endgroup$ – Austin Jan 15 '18 at 1:32