Can an Inverse Laplace transform have 2 different answers?

Note that $$\sinh 3t = \frac{e^{3t}-e^{-3t}}{2}$$ Hence, $$2e^t \sinh 3t = e^{4t} - e^{-2t}$$

No, both the Laplace and the inverse transform is unique.

You didn't follow through in using that you matched the transform of $\sinh bt$ you have that by definition $e^t 2\sinh 3t = e^t 2(e^{3t}-e^{-3t})/2 = e^{4t}-e^{-2t}$.