I would like to qualify the answers offered by @Bort and @sammy gerbil based on crude estimates given below.
Consider a fridge with perfect thermal insulation. Initially it is filled with air at room temperature, $T_{room}$. Let $V_{fridge}$ be interior volume of fridge. If all the air inside the fridge is to be cooled down to some average temperature $T_{cold}<T_{room}$, then you must remove heat
\begin{align}
Q=\rho_{air}V_{fridge}C_{p,air}(T_{room}-T_{cold})
\end{align}
Since fridge has perfect thermal insulation there is no loss, and $T_{cold}$ can be sustained indefinitely if you don't open the fridge. If you open the fridge $n$ times, then assuming the worst case scenario in that all the cold air inside the fridge is replaced by room air, you need to extract $(n+1)Q$ amount of heat.
Now consider another identical fridge, in which fraction of fridge volume, call it $\phi$, is replaced by water, initially everything at room temperature. Cooling down to $T_{cold}$ requires heat removal,
\begin{align}
Q'=[\rho_{water}~\phi V_{fridge}C_{p,water}+\rho_{air}(1-\phi)V_{fridge}C_{p,air}](T_{room}-T_{cold})
\end{align}
If you open fridge $n$ times as before, assuming again that all air inside is replaced by room air while also assuming that temperature of water remains unchanged and equal to $T_{cold}$ always, we need to remove heat equal to $Q'+n(1-\phi)Q$.
So the ratio of the amount of heat to be removed in the two cases is
\begin{align}
\alpha\equiv\frac{Q'+n(1-\phi)Q}{(n+1)Q}
\end{align}
Let $n$ be the average number of times you open the fridge between replacement of water inside the fridge.
For any given $\phi$, savings will result if
\begin{align}
\alpha & < 1 \\
\frac{Q'+n(1-\phi)Q}{(n+1)Q} & <1 \\
\frac{Q'}{Q}+n(1-\phi) & <(n+1) \\
\frac{Q'}{Q} & < 1+n\phi \\
n & >\frac{1}{\phi}(\frac{Q'}{Q}-1)
\end{align}
Since $\rho_{water}C_{p,water}\gg\rho_{air}C_{p,air}$, then if $1\geq \phi \gg 0$ we must have $\frac{Q'}{Q}\approx\frac{\rho_{water}C_{p,water}}{\rho_{air}C_{p,air}}\gg1$. Thus we need
\begin{align}
n>\frac{1}{\phi}\frac{\rho_{water}C_{p,water}}{\rho_{air}C_{p,air}}=\frac{1}{\phi}\frac{10^3\times 4.2\times 10^3}{1.2\times 1.0\times 10^3}\sim \frac{1}{\phi}10^3
\end{align}
Therefore even in the best case scenario where $\phi\approx 1$, you need to keep the water inside the fridge long enough that you open the fridge thousands of times between replacements of water, for any power savings to occur. This happens because the initial cost of cooling down water is much greater than that for air.
