350v @ 110 mA = 38.5 W and 12 V @ 1 A = 12 W, since 12 W is less than 38.5 W ...
That is the calculation for the power in the load not in the SSR.
The power dissipated in the SSR is given by \$ P_{SSR} = V_{SSR}I_{SSR} \$. Obviously when \$I = 0\$ then \$ P = 0 \$. When the SSR is on the power can be calculated by \$ P = I^2 R \$ and from the datasheet we can see that \$ R_{typ} = 18\ \Omega\$ so \$ P = 0.1^2 \cdot 18 = 0.18\ \text W\$.
Back to your misunderstanding: the internal wiring and semiconductor cross-sectional area, etc., determine the maximum current that can flow without exceeding the maximum current density (current per unit cross-sectional area) of the device. Operating beyond these limits will cause failure of some part of the device due to thermal effects. Your proposal was to exceed the current ratings by a factor of almost ten. This would be unlikely to last long.