I would say the quoted statement is ambiguous and possibly misleading.
Heteroskedasticity does not affect forecasting but serial correlation would
make point forecast invalid.
In general, forecast implications of residual diagnostics are:
No heteroskedasticity and no serial correlation
Forecast can be computed using consistent parameter estimates and forecast/prediction intervals have the right coverage probability.
Heteroskedastic but no serial correlation
Forecast can be computed using consistent parameter estimates. Forecast/prediction intervals would have the right coverage probability if sample size is large or if robust standard error is used.
Serially correlated
Parameter estimates are no longer consistent. Forecast and prediction intervals cannot be computed.
For example, take the simplest time series data generating process, the AR(1) model
$$
x_t = \rho x_{t-1} + \epsilon_t,
$$
and consider the following 3 cases.
Case 1: $\epsilon_t \stackrel{i.i.d.}{\sim} (0, \sigma^2)$
This is the ideal scenario.
The residual from fitting the AR(1) model to a sample would not have serial correlation, heteroskedasticity, or thick tails, because the population error term
$\epsilon_t$ does not.
The oracle one-period ahead forecast and mean-square forecast error (MSFE) are
\begin{align}
E[x_{t+1}|x_t] &= \rho x_t,\\
E[ (x_{t+1} - E[x_{t+1}|x_t])^2 ]&= \sigma^2.
\end{align}
So to compute one-period ahead forecast based on a sample of size $T$, you simply replace $\rho$ by, say, the OLS/conditional MLE estimate $\hat{\rho}$:
$$
x_{T+1 \vert T} = \hat{\rho} x_T.
$$
Same for the forecast mean square error
$$
\widehat{MSFE}^2 = \frac{1}{T} \hat{\sigma}^2 + \hat{\sigma}^2,
$$
where $\hat{\sigma}^2$ is the usual sum of squared residuals divided by $T-1$.
The 95% prediction interval is then $x_{T+1 \vert T} \pm 1.96 \times \widehat{MSFE}$.
This coverage probability of this prediction interval approaches the nominal coverage probability of 95% in large sample.
($\widehat{MSFE}$ can be computed as follows:
\begin{align}
\widehat{MSFE}^2 &= E[ (x_{t+1} - \hat{\rho} x_t)^2] \\
&= E[(\hat{\rho} - \rho)^2 x_T^2] + \sigma^2 \\
&\approx \frac{1}{T} \hat{\sigma}^2 + \hat{\sigma}^2.
\end{align}
In comparison with the oracle MSFE, the first term accounts for estimation error $\hat{\rho} - \rho$.
)
Case 2: $(\epsilon_t)$ is (conditionally) heteroskedastic but serially uncorrelated
(For example, $( \epsilon_t )$ could follow an ARCH process.
The consistency of $\hat{\rho}$ holds beyond such parametric specifications.)
The residuals from fitting the AR(1) model to a sample would show heteroskedasticity but no serial correlation.
The estimate $\hat{\rho}$ is still consistent, and the one-period ahead forecast is still $\hat{\rho} x_T$.
A prediction interval of the form $\hat{\rho} x_T \pm \cdots$ would still be correctly centered.
For the mean square forecast error,
$$
E[(\hat{\rho} - \rho)^2 x_T^2] \approx \frac{1}{T} \hat{\sigma}^2
$$
is no longer a good approximation. $\hat{\sigma}$ should be replaced by a heteroskedascitity-robust standard error. However, if $T$ is large, this term is negligible, and
$$
\hat{\rho} x_T \pm 1.96 \times \hat{\sigma}
$$
would still have asymptotic coverage probability of 95%.
Case 3: $(\epsilon_t)$ is serially correlated
(For example, $( \epsilon_t )$ could be itself AR(1).)
The residual from fitting the AR(1) model to a sample would have serial correlation.
The estimate $\hat{\rho}$ is no longer consistent (you can check this via simple simulation) and $\hat{\rho} x_T$ is no longer a consistent estimator of $E[x_{T+1}|x_T]$.
The minimal condition required for $\hat{\rho}$ to be consistent is
$\frac{1}{T} \sum_{t=1}^T E[x_t \epsilon_t] \rightarrow 0$. This would not be satisfied if $(\epsilon_t)$ has serial correlation.
Caveat: Best Forecast vs. Best Linear Forecast
Forecasting can be discussed in terms of the best forecast $E[x_{T+1}|x_T]$, or
best linear forecast.
The above discussion is in the context of the best forecast $E[x_{T+1}|x_T]$ (conditional mean of $x_{T+1}$ conditional on $x_T$).
In terms of the best linear forecast, the point forecast $\hat{\rho} x_T$ is still valid under Case 3.
The difference is that while $\hat{\rho}$ no longer consistently estimates $\rho$,
it still captures linear correlation between $x_{T}$ and $x_{T+1}$:
$$
\hat{\rho} \stackrel{p}{\rightarrow} \frac{Cov(x_{t+1}, x_t)}{Var(x_t)} \, (\neq \rho).
$$
The forecast interval
$$
\hat{\rho} x_T \pm 1.96 \times \hat{\sigma}_{HAC}
$$
would have the correct asymptotic coverage probability (with respect to the best linear forecast, not the best forecast) if $\hat{\sigma}^2_{HAC}$ is the heteroskedasticity autocorrelation robust (HAC) estimate of long-run variance computed from the residuals.