The key here is linearity; a (real-)linear map $\Phi$ defined on all self-adjoint bounded operators on $H$ has a unique extension to all of $B(H)$. This is due to the fact that every $B\in B(H)$ can be written as the linear combination of two self-adjoint operators via
$$
B=\Big(\frac{B+B^*}2\Big)+i\Big(\frac{B-B^*}{2i}\Big)\tag1
$$
so
\begin{align*}
\Phi':B(H)&\to B(H)\\
B&\mapsto \Phi\Big(\frac{B+B^*}2\Big)+i\,\Phi\Big(\frac{B-B^*}{2i}\Big)
\end{align*}
is well-defined, linear (straightforward computation), and an extension of the original map as $\Phi'\equiv\Phi$ on all self-adjoint operators.
For quantum states (i.e. the Schrödinger picture) there's a similar reasoning although one needs an extra step there:
Every (trace-class${}^1$) operator $A$ can be written as a linear combination of four positive semi-definite trace-class operators
As before this fact is what allows for a unique extension of a linear map defined on states to the full operator space. Finally, the:
Proof. First decompose $A$ into self-adjoint (trace-class) operators as in (1), and then use that every self-adjoint (trace-class) operator $A'$ can be written as the difference of two positive semi-definite operators through its spectral decomposition:
$$
A'=\sum_j a_j|g_j\rangle\langle g_j|=\underbrace{\sum_{j:a_j\geq 0}a_j|g_j\rangle\langle g_j|}_{A'_+\geq0}-\underbrace{\sum_{j:a_j<0}|a_j|\,|g_j\rangle\langle g_j|}_{A'_-\geq0}
$$
The final step is to turn these into states by normalizing them:
\begin{align*}
A&=\frac12\big(A+A^*\big)_+-\frac12\big(A+A^*\big)_--\frac i2\big(i(A-A^*)\Big)_++\frac i2\Big(i(A-A^*)\Big)_-\\
&=\tfrac{{\rm tr}((A+A^*)_+)}2\tfrac{(A+A^*)_+}{{\rm tr}((A+A^*)_+)}-\tfrac{{\rm tr}((A+A^*)_-)}2\tfrac{(A+A^*)_-}{{\rm tr}((A+A^*)_-)}\\
&\qquad\qquad\qquad-\tfrac{i{\rm tr}((i(A-A^*))_+)}2\tfrac{(i(A-A^*))_+}{{\rm tr}((i(A-A^*))_+)}+\tfrac{i{\rm tr}((i(A-A^*))_-)}2\tfrac{(i(A-A^*))_-}{{\rm tr}((i(A-A^*))_-)}\,.
\end{align*}
NB: of course if any of the ${\rm tr}((\ldots)_\pm)$ is zero then we could disregard them from the start; just to make sure we're not dividing by zero. $\square$
1: If you only care about finite-dimensional systems then you may replace "trace-class operator" by "matrix" because the trace class is an inherently infinite-dimensional concept which trivializes for finite dimensions.
