Although it may appear that the mean of the log-transformed variables is preferable (since this is how log-normal is typically parameterised), from a practical point of view, the log of the mean is typically much more useful.
This is particularly true when your model is not exactly correct, and to quote George Box: "All models are wrong, some are useful"
Suppose some quantity is log normally distributed, blood pressure say (I'm not a medic!), and we have two populations, men and women. One might hypothesise that the average blood pressure is higher in women than in men. This exactly corresponds to asking whether log of average blood pressure is higher in women than in men. It is not the same as asking whether the average of log blood pressure is higher in women that man.
Don't get confused by the text book parameterisation of a distribution - it doesn't have any "real" meaning. The log-normal distribution is parameterised by the mean of the log ($\mu_{\ln}$) because of mathematical convenience, but equally we could choose to parameterise it by its actual mean and variance
$\mu = e^{\mu_{\ln} + \sigma_{\ln}^2/2}$
$\sigma^2 = (e^{\sigma^2_{\ln}} -1)e^{2 \mu_{\ln} + \sigma_{\ln}^2}$
Obviously, doing so makes the algebra horribly complicated, but it still works and means the same thing.
Looking at the above formula, we can see an important difference between transforming the variables and transforming the mean. The log of the mean, $\ln(\mu)$, increases as $\sigma^2_{\ln}$ increases, while the mean of the log, $\mu_{\ln}$ doesn't.
This means that women could, on average, have higher blood pressure that men, even though the mean paramater of the log normal distribution ($\mu_{\ln}$) is the same, simply because the variance parameter is larger. This fact would get missed by a test that used log(Blood Pressure).
So far, we have assumed that blood pressure genuinly is log-normal. If the true distributions are not quite log normal, then transforming the data will (typically) make things even worse than above - since we won't quite know what our "mean" parameter actually means. I.e. we won't know those two equations for mean and variance I gave above are correct. Using those to transform back and forth will then introduce additional errors.
Best Answer
Short answer: Back-transformed coeffient estimator is biased, and not a good estimator. Back-transformed confidence interval is valid, but sub-optimal.
Longer answer: Since you have not included any residual plots in your post, it is unclear whether your transformed model actually fits the data well. (This is not something you can tell from the coefficient estimates table.) The OLS coefficient estimators in the Gaussian linear regression model are MLEs, so if you take the corresponding back-transformed estimators these will be MLEs of the corresponding back-transformed parameters (by the invariance properties of MLEs). However, you should bear in mind that back-transformed estimators using a non-linear transform will be biased, which is why many statisticians counsel against their use (see e.g., this related question). In the case of an exponential back-transformation you will get an estimator that is positively biased, meaning that it will tend to overestimate the true back-transformed parameter on average. The back-transformed coefficient estimator as not a particularly good point estimator, for this reason.
You can also back-transform the limits of the confidence intervals, and these remain valid interval estimators, with the same confidence level as in the original model. However, by using a non-linear back-transformation you will end up with a confidence interval that is a little wider than it needs to be (since the equal-tail interval used for the linear model is no longer the shortest interval with that confidence level). It is possible to get a shorter interval at the same confidence level in the space of the back-transformed parameter, but this requires a bit more work, and it requires more knowledge of the underlying properties of the pivotal quantity used to form the interval estimator.