返回任务时我不应该使用等待/异步吗？

Question

When I want to create a method returning Task with I/O bound work I usually use something like this:

public async Task<ProductModel> GetProduct(string name, CancellationToken ct = default)
{
    //...
    return await GetAsync<ProductModel>(request, ct);
}

And call it like this:

var product = await GetProduct("some product", ct);

But I came across some function definition just like above but without the await/async :

public Task<ProductModel> GetProduct(string name, CancellationToken ct = default)
{
    //...
    return GetAsync<ProductModel>(request, ct);
}

With a similar call as mine.

My question is, is this latter definition safe, can I just suppress the await/async in the function returning a Task ?

And if so, are there any benefits or drawbacks?