Do Microwave oven heating times grow linearly with Wattage? Calculating optimal heating time
The rate at which a mass absorbs microwave radiation is characterized by the 'Specific Absorption Rate', which is proportional to the electromagnetic field intensity:
Wikipedia has a dedicated article to this phenomenon but in short it says
$$\text{SAR} = \int_\text{sample} \frac{\sigma(\vec{r})|\vec{E}(\vec{r})|^2}{\rho(\vec{r})} d\vec{r}$$
Because the absorption rate is proportional to the EM field intensity, $|\vec{E}(\vec{r})|^2$, which is in turn proportional to power, then the relationship will indeed be linear.
Assuming 100% energy efficiency (which is a wild overestimate: 20% might be more accurate but I do not know the answer to that question) your total 'energy' transferred to your dinner will be:
$$ \text{Energy} = \text{Power} \Delta t$$
i.e.
$$\Delta t = \frac{\text{Energy}}{\text{Power}} $$
The cook time will be inversely proportional to your oven power.
1100 watts for 3.5 minutes computes to $$ \text{Energy } = 1100 \text{ Watts } \times 210 \text{ seconds } = 231,000 \text{ Joules}$$
700 Watts for 5.5 minutes computes to $$ \text{Energy } = 700 \text{ Watts } \times 330 \text{ seconds } = 231,000 \text{ Joules}$$
Thus a 900 Watt oven would necessitate
$$\Delta t = \frac{\text{Energy}}{\text{Power}} = 256.66 \text{ seconds } = 4.278 \text{ minutes }$$
To a first approximation, the product of power $P$ and time $t$ should be constant, since the energy put into heating the object is $$ E = Pt. $$ That is, to achieve the same effect, time should be inversely proportional to power. (Of course, if the power is so low that the process lasts for hours, one would need to consider the rate at which the food is cooling as well.)
Note that $$ (1100\ \mathrm{W}) \times (210\ \mathrm{s}) = 231\ \mathrm{kJ}, $$ while $$ (700\ \mathrm{W}) \times (330\ \mathrm{s}) = 231\ \mathrm{kJ}, $$ so our assumption of constant energy seems to be what the manufacturer used. Then with a $900\ \mathrm{W}$ power source the required time is $$ t = \frac{E}{P} = \frac{231\ \mathrm{kJ}}{900\ \mathrm{W}} = 257\ \mathrm{s} $$ total, to the nearest second. Note that this is not quite the mean of $210\ \mathrm{s}$ and $330\ \mathrm{s}$.
By analogy, think of this in terms of distances (rather than energies), speeds (rather than powers), and times. If a destination is $12$ miles away, you can walk at $1\ \mathrm{mph}$ for $12$ hours, $3\ \mathrm{mph}$ for $4$ hours, or $2\ \mathrm{mph}$ for $6$ (but not $(12+4)/2 = 8$) hours.