Yes. It's like dumping floating mirrors into the ocean.

• The white of ice cubes reflects light (read: heat energy)
• Part of the reflected heat energy would otherwise be absorbed by the dark blue sea
• Ice reduces the amount of energy coming into earth
• You don't have to continuously dump ice cubes
• Thermodynamics have not been violated

If you want to extend the effect, dump large ice cubes into the ocean - less surface area means more not-melted time.

For every $O(n)$ ice cubes dumped, you get $O(n^2)$ cooling effect, because ice cubes don't melt at $O(1)$ pace*. They melt at $O(\frac{1}{n})$ pace*.

For all those who aren't computer scientists, $O(...)$ notation represents asymptotic pace;

$O(1) = 3$

$O(n) = 6x+6$

$O(n^2) = 8x^2+350x+5$

_{*per cube. Really, they don't melt at $O(n)$ pace per $cm^3$ of ice, but $O(n)$ pace per $cm^2$ of surface area. The surface area of a $O(n)\space cm^3$ cube can be informally approximated to $O(\frac{1}{n})\space cm^2$.}

An extremely simple explanation

$h$ - heat in ocean

$i$ - heat coming in per second

We take freshwater and freeze it, dumping the heat into the ocean.

$h$ goes up by $W$ every time we put 1 cube in. $i$ goes down by $G$ for $H$ seconds every time we put 1 cube in.

We just need $G*H > W$.

And don't forget that in reality $H$ is not constant, it goes down per the size of the cube.

$Find\space cube\space size \space s \space such \space that\space G*h(s) > W.$

As if it wasn't obvious enough,

Then continue dumping floating mirrors.