All the existing answers nicely explain the issue, but I'll add some numbers to the discussion:
In 2018, 112 objects have been launched to orbit. Let's assume a typical rocket mass of 500 t and a payload of 10 t (values for the Falcon 9). Of these 112 objects, 30 were launched into high orbits where part of the rocket exhaust may have escaped Earth's sphere of influence (SOI), and four (Elon Musk's Tesla Roadster, InSight, Parker Solar Probe, and BepiColombo) were launched into interplanetary orbits where they and their upper rocket stages, escaped Earth's SOI themselves.
For high orbits, we can assume that about half of the payload's mass leaves Earth's SOI in the form of rocket exhaust. For payloads that are launched into interplanetary space, we can assume that an additional 10 tons are sent in the form of the launcher's spent upper stage.
Assuming that 2018 is a representative year and those guesses are about correct (variations from 0.1 to 10 would be possible),
- The total launched mass is $5 \cdot 10^7\ kg/year$
- The total satellite mass increases by $1 \cdot 10^6\ kg/year$
- The total satellite mass in high orbits increases by $3 \cdot 10^5\ kg/year$
- The total mass leaving Earth's SOI, from both interplanetary launches and rocket exhaust of high-orbit satellites, is $3 \cdot 10^5\ kg/year$
As discussed in the other answers, the mass of satellites that stay inside Earth's SOI has no influence on the distance between Earth and the Sun. The only relevant part is the $3 \cdot 10^5\ kg/year$ that are launched away from Earth.
As you correctly noted, the mass itself isn't relevant, the impulse is, so let's assume that the mass leaves Earth at 10 km/s. The fastest-ever spacecraft to leave Earth was New Horizons at 16 km/s, while a typical Mars transfer takes 6 km/s. Rocket exhaust is typically much slower, on the order of 3 km/s.
We arrive at an impulse of $ 3 \cdot 10^9\ Ns / year $, or an average force of $ 100\ N $.
But in which direction is the force applied? For interplanetary probes the ejection direction depends on your intended destination and transfer plan, while for rocket exhaust from satellites in high orbits it's pretty much random. All in all, I guess that it will pretty much even out. After all, in 2018 two launches were to higher solar orbits and two were to lower solar orbits.
If the force is applied in the same direction in which Earth is moving, it is accelerated and thus lifted into a higher orbit. If the force is applied in the opposite direction, it is decelerated and moves into a lower orbit (closer to the Sun).
The orbital energy of Earth is calculated as $ -\frac{ G \cdot M \cdot m_{Earth}}{2a}$, where G is the constant of gravity, M is the total mass of the Earth-Sun system, and a is the semi-major axis (average distance between Earth and Sun).
The power at which Earth is accelerated or decelerated is $\frac{dE}{dt} = Fv$, where F is the force of acceleration and v is the orbital velocity of Earth. We arrive at
$a(t) = - \frac{G \cdot M \cdot m_{Earth}}{2 \cdot (E_0 \pm F v t)} = - \frac{G \cdot M \cdot m_{Earth}}{2 \cdot E_0} \cdot \frac{1}{1 \pm \cfrac{F v}{E_0} \cdot t} $
$\frac{da(t)}{dt} = - \frac{G \cdot M \cdot m_{Earth}}{2 \cdot E_0} \cdot \frac{da(t)}{dt} (\frac{1}{1 \pm \frac{F v}{E_0} \cdot t}) = - \frac{G \cdot M \cdot m_{Earth}}{2 \cdot E_0} \cdot \frac{\mp \frac{F v}{E_0}}{(1 \pm \frac{F v}{E_0} \cdot t)^2} $.
$\frac{F v}{E_0} \cdot t$ is negligible compared to 1 (about $t \cdot 3 \cdot 10^{-20} / year$), so the formula is simplified to:
$\frac{da}{dt} = \pm \frac{G \cdot M \cdot m_{Earth}}{2 \cdot E_0} \cdot \frac{F v}{E_0} = \pm GM \cdot \frac{m_{Earth}\cdot v}{2 \cdot {E_0}^2} \cdot F = 1.8 \cdot 10^{-18}\ \frac{\frac{m}{s}}{N} \cdot F$.
So the result on Earth's distance from the Sun is $\pm 1.8 \cdot 10^{-16}\ m/s = \pm 5 \cdot 10^{-9}\ m/year$, assuming that everything is ejected in the same direction.