Q1: Yes. You ask ``If $X$ is a countable dimensional dense subspace of the Banach space $Y$, are the operators on $Y$ which leave $X$ invariant dense in the operators on $Y$?" Use Mackey's argument for producing quasi-complements (just a biorthogonalization procedure, going back and forth between a space and its dual) to construct a fundamental and total biorthogonal sequence $(x_n,x_n^*)$ for $Y$ with the $x_n$ in $X$; even a Hamel basis for $X$. Now use the principle of small perturbations to perturb an operator on $Y$ to a nearby one that maps each $x_n$ back into $X$. I am traveling now and so can't provide details or references, but I think that is enough for you, Yemon. The key point is that the biorthogonality makes the perturbation work--if $x_n$ were only a Hamel basis for $X$ it is hard to keep control.
I have my doubts whether this result appears in print even if oldtimers like me know the result as soon as the question is asked.
EDIT 7/4/10: Once you get the biorthogonal sequence $(x_n,x_n^*)$ with $x_n$ a Hamel basis for $X$, you finish as follows: WLOG $\|T\|=1$ and normalize the BO sequence s.t. $\|x_n^*\|=1$. Define the operator $S$ on $X$, the linear span of $x_n$, by $Sx_n=y_n$, where $y_n$ is any vector in $X$ s.t. $\|y_n-Tx_n\| < (2^{n}\|x_n\|)^{-1}\epsilon$. On $X$ you have the inequality $\|T-S\|<\epsilon$, so you get an extension of $S$ to $Y$ that satisfies the same estimate on $Y$. In checking the estimate you use the inequality $\|x\| \ge \sup_n |x_n^*(x)|$; i.e., biorthogonality is crucial.
To get the biorthogonal sequence, you take any Hamel basis $w_n$ for $X$ and construct the biorthogonal sequence by recursion so that for all $n$, span $(w_k)_{k=1}^n = $ span $(x_k)_{k=1}^n$. At step $n$ you choose any $x_n$ in span $(w_k)_{k=1}^n $ intersected with the intersection of the kernels of $x_k^*$, $1\le k < n$, and use Hahn-Banach to get $x_n^*$.
The Mackey argument I mentioned gives more. If you have any sequence $w_n$ with dense span in $Y$ and any $w_n^*$ total in $Y^*$, with a back and forth biorthogonalization argument you can build a biorthogonal sequence $(x_n,x_n^*)$ s.t. for all $n$, span $(x_k)_{k=1}^{2n}$ contains span $(w_k)_{k=1}^n $ and span $(x_k^*)_{k=1}^{2n}$ contains span $(w_k^*)_{k=1}^n $. This is quite useful when dealing with spaces that fail the approximation property; see e.g. volume one of Lindenstrauss-Tzafriri and, for something recent, my papers with Bentuo Zheng, which you can download from my home page.
EDIT 7/11/10: Getting a general positive answer to Q2 would be very difficult. Although not known to exist, it is widely believed that there is a Banach space with unconditional basis upon which every bounded linear operator is the sum of a diagonal operator and a compact operator. On such a space, the operators that map $\ell_1$ into itself would be dense in the space of all bounded linear operators.
Best Answer
In
Argyros, Spiros A.; Arvanitakis, Alexander D.; Tolias, Andreas G. Saturated extensions, the attractors method and hereditarily James tree spaces. Methods in Banach space theory, 1–90, London Math. Soc. Lecture Note Ser., 337, Cambridge Univ. Press, Cambridge, 2006
the authors construct a separable space $Z$ so that $X:= Z^*$ is non separable, hereditarily indecomposable (HI), and every strictly singular operator on $X$ is weakly compact and hence has separable range. Of course, $X$ embeds isometrically into $\ell_\infty$ because $Z$, being separable, is a quotient of $\ell_1$. There is no operator from this $X$ into $X \oplus X$ that has dense range. Indeed, if $T$ is such an operator, then letting $P_i$ for $i= 1,2$ be the projections onto the factors of $X \oplus X$, we see that $P_iT$ is of the form $\lambda_i I + W_i$ with $W_i$ strictly singular by the usual Gowers-Maurey HI theory. Both operators $P_iT$ have dense range, so neither $\lambda_i$ is zero because both $W_i$ have separable range. To see that $T$ cannot have dense range, observe that $T^*$ is not injective. Indeed, take a norm one $x^*$ in $X^*$ which vanishes on $W_1X+W_2X$. Then $T^* (\lambda_1^{-1}x^*, - \lambda_2^{-1}x^*)= 0$.
There is also no injective dense range operator from $X\oplus X$ into $X$. For suppose that $S(x_1,x_2) = S_1 x_1 + S_2 x_2$ is such an operator. Write $S_i= \lambda_i I + W_i$ with $W_i$ strictly singular. Since the $W_i$ have separable range and $S$ has dense range, one of the $\lambda_i$, say $\lambda_1$, is not zero, and hence $S_1$ is Fredholm of index zero. If $S_2$ has finite rank, then clearly $S$ is not injective, but otherwise the infinite dimensional $S_2X$ intersects nontrivially the finite codimensional $S_1X$, which again implies that $S$ is not injective.