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It would. !!! The 1/2 factor doesn't change Lorentz invariance of the metric measure, but it's quite essential to the formalism that comes later. One would think we're done with negative energies/frequencies, and such. But no. They keep biting our buttocks later with the Fourier transform. That's where the Stueckelberg-Feynman prescription for antiparticles comes in.1 point
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Just to answer the OP question, It would not. Without the step function it would be \[\int dk^0 \delta (k^2-m^2) =\frac 1 {\omega_k} \] rather than \(\frac 1 {2 \omega_k}\).1 point