BIBO Stability considerations for Transfer Functions
If \$G(s)\$ is an arbitrary transfer function it is BIBO stable if and only if it is linearly stable. The proof is simple. Let \$x(t)\$ be a bounded input and put \$x_0\$ as the least-upper-bound of \$x(t)\$. The Laplace transform gives

$$\frac{x_0}{s} &gt; X(s)$$

hence

$$G(s)\frac{x_0}{s} &gt; G(s)X(s)$$

hence 

$$\mathscr{L}^{-1} \left\{G(s)\frac{x_0}{s} \right\} &gt; \mathscr{L}^{-1}\{G(s)X(s)\}.$$

But since \$G(s)\$ is stable, \$\frac{G(s)}{s}\$ is also at least quasi-stable (has poles no further than the origin), so \$\mathscr{L}^{-1}\left\{G(s)\frac{x_0}{s} \right\}\$ is also bounded.
Indeed \$\mathscr{L}^{-1} \{G(s)X(s)\} \$ will be a linear combination of terms within a collection of exponentially-decreasing envelopes, so the BIBO-linear stability correspondence is actually quite intuitive. 

Pole cancellation can be done while maintaining stability, but it's risky because closed-loop poles move from their open-loop position and modeling / control uncertainties can cause the zeros to move off of the poles. If you use pole-zero cancellation to delete unstable poles the actual positions of the poles and zeros may not quite align, and the response will be unstable. SO NEVER DO THIS!!! The initial energy does not matter, as your inputs will excite modes corresponding to every pole which is not completely cancelled. The recommended approach for unstable poles, as always, is to use feedback to stabilize them.