The route to the answer is somewhat anti-intuitive. By reflecting some of the Sun's energy back towards the sun at a point you are effectively reducing the flux of energy that can emerge from the photosphere and escape.
The global effect of this on the Sun must be similar to that of blocking the flux at the photosphere - in other words, similar to the effects of sunspots. The local effect on the temperature structure will of course be completely different, because sunspots are places where the photospheric temperature is much (1000 K) cooler than the unspotted photosphere. Here, you would be creating a hotspot, nevertheless, the flux emerging from the surface and escaping to infinity would be lower than it would for a star of the same radius and effective temperature where there was no mirror.
The local effects really would be quite local. Convective energy transfer is very effective just below the photosphere, so the excess energy is redistributed on a local convective turnover timescale (five minutes).
The global effects can be treated in a similar way to the effects of sunspots. The canonical paper on this is by Spruit & Weiss (1986). They show that the effects have a short term character and then a long term nature. The division point is the thermal timescale of the convective envelope, which is of order $10^{5}$ years for the Sun.
On short timescales the nuclear luminosity of the Sun is unchanged, there will be an additive effect due to the hot spot on the surface, but the stellar structure remains the same as does the surface temperature. As about half the flux from the hotspot goes into the Sun and only half goes into space, the net luminosity at infinity (after subtracting that blocked by the mirror) will be lower, whilst the flux at the mirror will increase.
On longer timescales, the luminosity will tend to stay the same because the nuclear burning core is unaffected by what is going on in the thin convective envelope. However, roughly half the flux reflected by the mirror can't escape from the star. To lose the same luminosity it turns out that the radius increases and the photspheric area not affected by the reflected beam (the "unspotted region") gets a little hotter. In this case, the radius squared times the photospheric temperature will increase to make sure that the luminosity observed beyond the mirror stays the same - i.e. by $R^2T^4(1 - \beta) = R_{\odot}^2 T_{\odot}^4$, where $\beta$ is the fraction of the solar luminosity intercepted by the mirror.
The calculations of Spruit et al. (1986) indicate that for $\beta=0.1$ the surface temperature increases by just 1.4% whilst the radius increases by 2%. Thus $R^2 T^4$ is increased by a factor 1.09. This is not quite $(1-\beta)^{-1}$ because the luminosity does drop slightly.
So yes, if you keep the mirror there for longer than $10^5$ years you will increase the temperature of the Sun, but perhaps not by as much as you would have thought.
Further edit:
The above discussion is true for the Sun because it has a very thin convection zone and the conditions in the core are not very affected by conditions at the surface. As the convection zone thickens (for example in a main sequence star of lower mass), the response is different. The increase in radius becomes more pronounced; to maintain hydrostatic equilibrium the core temperature decreases and hence so does the nuclear energy generation. The luminosity of the star falls and the surface temperature stays roughly the same.
This is why I have made comments on other answers here, because although they correctly state that the Sun will get hotter, it is not obvious that this should be so and indeed would not be so for a lower mass star.
