EDIT: The OP added some questions, including one about the case of a right half-plane cohomological bicomplex $(B^{i,j}, d_h, d_v)$, where $B^{i,j} = 0$ for $i<0$, $d_h : B^{i,j} \to B^{i+1,j}$ and $d_v : B^{i,j} \to B^{i,j+1}$. Suppose that $Z^{i,j} \subset B^{i,j}$ is such that $d_h(Z^{i,j}) = 0$, and filter the total complex $$ (C, d) = (\bigoplus_{i,j} B^{i,j}, d_h + d_v) $$ by $$ F^s = \bigoplus_{j-i>s} B^{i,j} \oplus \bigoplus_{j-i=s} Z^{i,j} $$ for all integers $s$. Here $C$ and $F^s$ are graded, with $B^{i,j}$ and $Z^{i,j}$ in degree $i+j$. Then $(F^s, d)$ is a subcomplex of $(C, d)$, and contains $(F^{s+1}, d)$ as a further subcomplex. We get an exact couple in the usual way, with $A^{s,t}$ and $E_1^{s,t}$ equal to the degree $s+t$ parts of $H(F^s, d)$ and $H(F^s/F^{s+1}, d)$, respectively.
I claim that $\lim_s F^s = 0$ and $\lim^1_s F^s = 0$. This can be checked one degree at a time, since $F^s = 0$ in degrees less than $s$. It follows (see Boardman's Theorem 9.2) that $\lim_s A^s = 0$ and $\lim^1_s A^s = 0$, so the spectral sequence is conditionally convergent to the colimit $G = H(C, d)$.
Furthermore, $F^s/F^{s+1}$ and $E^1_s$ are concentrated in degrees $\ge s$, corresponding to $t\ge0$, so this is an upper half-plane cohomological spectral sequence with exciting differentials. Hence it is strongly convergent to $G$, by the modified form of Boardman's Theorem 6.1 that I mentioned above.
To prove the modified form, one does as Boardman says. Let $F^s G$ be the image of $H(F^s, d)$ in $G$. One must check that the filtration $\{F^s G\}_s$ of $G$ is complete Hausdorff and exhaustive, and that the natural inclusion $F^s G/F^{s+1} G \to E_\infty^s$ is an isomorphism. Both claims can be checked one degree at a time, and for each degree the proof of Theorem 6.1(a) carries over. (I do not have Cartan--Eilenberg at hand: I do not recall if they spelled this out.)