You want a very basic AM transmitter. The old Knight-Kit C100 walkie-talkie has a very simple design. It used a super-regenerative receiver, to minimize parts count. It had one crystal, to set the transmitting frequency, and three transistors. You could probably simplify that, to make just a transmitter. The schematic USED to be available on the Web, but it appears to have vanished.
OLD copies of the Radio Amateur's Handbook should have circuits that could be adapted to a simple transistor AM transmitter.
If I were going to try to throw something together, I'd probably go DSB rather than AM, and use an SA612AN (NE602 replacemement). Instead of a VFO, I'd hang the crystal on the SA612AN, and I'd probably try a plastic 2N2222A, either barefoot or driving a 2N3866. I'm not sure what I'd use for an audio preamp. Here's a more elaborate transceiver, that illustrates the basic principles.
Now for the bad news. There's no way you're going to power this from a CR2032 coin battery. Look at camera batteries, like a Kodak 2CR5 6V lithium battery.
Or will there be some modifications? For example getting triangular a bit?
No, not if the filter is linear. By more or less definition, the output signal from a linear filter does not contain any frequency that isn't present in the input signal.
If the input is a sinusoidal signal and the output is not a sinusoidal signal, even if by just a bit, the filter is not linear since, as Fourier analysis shows, a non-sinusoidal signal necessarily has multiple sinusoidal components of different, related frequencies.
Thus, to make the sinusoid triangular a bit requires adding frequency components that are not present in the input signal, i.e., adding harmonic distortion.
In summary, if the filter is linear, a sinusoidal input of (angular) frequency \$\omega\$ will result in a sinusoidal output of frequency \$\omega \$ with, at most, a modified amplitude and phase.
$$v_I(t) = \cos\omega t $$
$$v_O(t) = |H(\omega)|\cos[\omega t + \phi(\omega)] $$
Best Answer
a) If your signal is AM-DSB, the original carrier power is \$P\$ and "cutting off" the one sideband means cancelling it out completely, then the calculations are as following: A 20dB amplification equals a power amplification of 100. Both the left and unattenuated sideband and the reduced carrier get amplified.
\$\frac{P}{4}\cdot 100+\frac{P}{100}\cdot 100=200W\$
\$26P=200W\$
\$P\approx7.692W\$
b) However, if it was an AM-SSB signal and you cancelled out the only one sideband, your calculations are correct.