Does an expanding event horizon "swallow" nearby objects?

A quick introductory note: from the comments it's clear the OP is thinking of a black hole merger. My answer was written before I realised this so it assumes the black hole is growing outwards by the gradual accretion of matter. My argument wouldn't be applicable to black hole mergers.

There isn't an analytic solution for the situation you describe, but there is a related model system that we use to get an idea of what happens. This is the Oppenheimer-Snyder metric. The OS metric describes a sphere of uniform density, pressureless gas collapsing under its own gravity. Real stars are neither uniform nor pressureless so the OS metric can at best give us a guide as to the main features of the collapse, but let's go with it and see what happens.

In the rest frame of an observer on the surface of the collapsing ball the event horizon first appears at the centre of the sphere and grows outwards towards the observer as the ball collapses. The event horizon passes the observer at the moment when the radius of the sphere is equal to its Schwarzschild radius.

The relevance to your question is that we can consider the observer on the surface as your object hanging at the horizon. The collapse causes the horizon to grow past and engulf the observer as you describe in your question. The OS metric allows us to calculate the time at which this happens both in the rest frame of the infalling observer and in the frame of the observer far from the collapsing sphere.

The equations we need are given in this article on the GR Wiki. We use a timelike parameter $T$ - note that this is not the proper time of any observer, just a parameter. Then the proper time of the infalling observer is given by:

$$ t' = \frac{A(0)}{2}\left(T + \sin T\right) \tag{1} $$

The parameter $A$ is the scale factor describing the collapse of the sphere i.e. $A$ starts at a finite value at the beginning of the collapse and decreases to zero at the moment the singularity forms. $A$ is given by:

$$ A(T) = \frac{A(0)}{2}\left(1 + \cos T\right) \tag{2} $$

and $A(0)$ is related to the initial radius of the sphere $r_0$ as measured by the observer far from the black hole by:

$$ A(0) = \sqrt{\frac{r_0{}^3}{r_s}} $$

The scale factor $A$ falls to zero when $T=\pi$, so collapse occurs over the range from $T=0$ to $T=\pi$. Substituting $T=\pi$ in equation (1) gives a finite answer so the collapse completes in a finite time as measured by our observer sitting on the surface of the sphere. That means:

The observer on the surface of the sphere observes themselves to pass the growing event horizon in a finite time

The question asks about the view of a remote observer. If we take the observer out to infinity then the observer's coordinates are the Schwarzschild coordinates and specifically the observer's time is the Schwarzschild $t$ coordinate. For this observer the time at the surface of the sphere, i.e. at the position of the infalling observer, is given by the rather ugly expression:

$$ t = r_s\left( \ln\left( \frac{\sqrt{r_0/r_s - 1} + \tan(T/2)}{\sqrt{r_0/r_s - 1} - \tan(T/2)} \right) + \sqrt{r_0/r_s - 1}\left( T + \frac{r_0}{2r_s}(T + \sin T)\right) \right) \tag{3} $$

Although this is rather involved we need only note that the right hand side goes to infinity when the denominator in the log term goes to zero i.e. when:

$$ \sqrt{\frac{r_0}{r_s} - 1} = \tan\left(\frac{T}{2}\right) \tag{4} $$

and this is the moment when the radius of the sphere is equal to the Schwarzschild radius (as measured by the observer at infinity). So the conclusion is:

The observer far from the sphere observes the observer on the surface of the sphere to take an infinite time to pass the point $r=r_s$

So if you are prepared to accept the above as an acceptable model for the situation you describe then neither of the two options you present is correct. For the distant observer no event horizon ever forms and the the infalling observer takes an infinite time to pass the point $r = r_s$ where the horizon would form given infinite time.

I would guess you are thinking of an established black hole with a horizon at $r=r_s$, and what happens if this horizon grows (maybe because a load of mass is dumped into the black hole). The problem is that this is an unphysical situation as for the distant observer an event horizon takes infinite time to form. So the experiment could never be done. The calculation I've described (given the limitations of the OS metric) illustrates what would actually happen.

As it is usually defined, the event horizon of a black hole is the imaginary surface which nothing can escape to infinity from. You cannot actually identify the event horizon of a black hole without taking into account everything that will happen in the indefinite future. So if you add more mass to a black hole at time $t$, the position of the horizon at all times previous to $t$ gets recalculated, and expands slightly. This recalculation alters the position at times much less than $t$ infinitesimally, though; it changes it just enough to let a ray of light moving outward at the horizon escape at infinite time.

Unlike a ray of light escaping, however, an object falling into the black hole will not move outward when the horizon expands due to new mass falling in. Its calculated position will retroactively move outward infinitesimally when the mass is added at time $t$, but it will still be well within the new horizon, assuming it is significantly larger than the old horizon.

Note that this analysis is classical, and does not take into account the quantum mechanics of black holes. If firewalls exist, and are taken into account, the answer could be completely different.