I don't know, but I would be very surprised if a topology made the problem any easier since I would imagine that you could always use the induced topology from $L(X)$ on $A$ so if you can answer it in the topological setting then you can answer it in the discrete setting. More specifically, the question is about the image of $T$ but imposing a topology on $A$ only messes around with the domain so I'd be surprised if it had a significant effect.
However I suspect that I haven't understood the problem very well since it seems as though a simple example would be where $A$ was the compact operators together with the unit. That seems to fit the conditions but it certainly isn't dense in the strong operator topology (though it is in the weak topology).
Edit: As I suspected, I hadn't. As Matthew points out, the topology is not the strong operator topology but the weak operator topology with respect to the strong topology on $X$. That is, $T_\gamma \to T$ if $T_\gamma x \to Tx$ for all $x$.
This, then, sounds a lot like the approximation question which is known to be false for arbitrary Banach spaces (paper of Eno, though I forget the exact reference) since in a space where the approximation property fails one could take the subalgebra of finite rank operators (plus the identity to make it unital).
However, my first point still seems valid.