Here are examples to show that neither of the two conditions 1. and 2. that you give as jointly sufficient can just be dropped.
Take $X$ to be two-dimensional real space with the pointwise order. Norm it by taking the closed unit ball to be the convex hull of
$\{(-3,3),(0,2),(2,0),(3,-3),(0,-2),(-2,0),(-3,3)\}$. On the positive cone, $\|(x,y)\|=(x+y)/2$.
Clearly, $(-1,-1)\le (-1,1)\le (1,1)$ and $\|(-1,-1)\|=\|(1,1)\|=1$ whilst $\|(-1,1)\|=1/3$.
Any attempt to write $(-1,1)=u-v$ with $u$ and $v$ positive will have to have $\|u\|,\|v\|\ge 1/2$, so your condition 1 fails. Condition 2 does hold.
Define $T(x,y)=(y,y)$, which is certainly positive. For $(x,y)\ge 0$, $\|T(x,y)\|=y\le 2\|(x,y)\|$.
On the other hand $\|T(-1,1)\|=\|(1,1)\|=1$ whilst $\|(-1,1)\|=1/3$ so that $\|T\|\ge 3$.
For the second example, take $Y$ to be two-dimensional real space with the pointwise order and normed by $\|(x,y)\|=\max\{\|(x,y)\|_\infty, |x-y|\}$.
Take $Z$ to be two-dimensional real space with the pointwise order and supremum norm. For $(x,y)\ge (0,0)$ the two norms are equal.
$Y$ has your property 1 but not 2. The identity map from $Z$ into $Y$ has norm at least 2 as $\|(-1,1)\|=2$ whilst $\|(-1,1)\|_\infty=1$.
On the positive cone all norms of images are equal. To get an example from a space into itself, take $X=Y\oplus Z$ and the operator
$(y,z)\mapsto (z,0)$.
Recently I found a counterexample to a somewhat stronger version of the question I originally asked, but which I actually needed. Namely for a smooth map $f\colon X\to Y$ which is transversal to $\Lambda\subset T^*Y\backslash 0$ (see e.g. Hormander's book for the definitions) the map $f^*\colon C^{-\infty}_{\Lambda}(Y)\to C^{-\infty}_{f^*\Lambda}(X)$ may not be topologically continuous (notice the subscript $f^*\Lambda$ in the target).
The counterexample is quite simple. Let $f\colon \mathbb{R}^2\to \mathbb{R}$ be the projection to the first coordinate. Let $\Lambda=\mathbb{R}\times (\mathbb{R}\backslash 0)$. Then $C^{-\infty}_{\Lambda}(\mathbb{R})=C^{-\infty}(\mathbb{R})$ is the usual Schwartz space with the usual weak topology. Furthermore $f^*\Lambda=\{(x,y;\xi,0)\}$. Then the map $f^*\colon C^{-\infty}_\Lambda(\mathbb{R})\to C^{-\infty}_{f^*\Lambda}(\mathbb{R}^2)$ is not topologically continuous.
Strictly speaking I do not yet have a counterexample to my original question, but this one already suffices my purposes.
Best Answer
A sufficient (additional) condition is that $B$ be compact for some Hausdorff vector topology for $V$. The proof goes as follows.
Letting $\langle x_i\rangle$ be a Cauchy sequence, it is contained in some $nB$, and so it has some cluster point $y$ there. Given $\varepsilon>0$, there is $i_0$ such that $x_i-x_j\in\frac 12\varepsilon B$ for $i,j\ge i_0$ , and it remains to show that $x_{i_0}-y\in\frac 12\varepsilon B$ . Indeed, if this does not hold, we have $y\not\in x_{i_0}-\frac 12\varepsilon B$ . Since $B$ is compact, it is closed in the Hausdorff case, and so is $x_{i_0}-\frac 12\varepsilon B$ as we have a vector topology. By the cluster point property, there must be some $i\ge i_0$ with $x_i\not\in x_{i_0}-\frac 12\varepsilon B$ , which is impossible.
Edit. Actually, I originally had in mind a more general condition, but I couldn't correctly recall it when writing the answer. Namely, a sufficient condition is that there be a Hausdorff topological vector space $E$ and there an absolutely convex compact set $C$ and a linear map $\ell:V\to E$ with $B=\ell^{-1}[C]$ , and such that we have $y\in{\rm rng\ }\ell$ whenever $y\in E$ is such that $(y+\varepsilon C)\cap({\rm rng\ }\ell)\not=\emptyset$ for all $\varepsilon>0$ . The proof is essentially the same as the one given above with $C$ in place of $B$ . This more general condition applies for example in the case where $B$ is the closed unit ball of $C^k([0,1])$ since we can take $E=({\mathbb R}^{[0,1]})^{k+1}$ and $C=([-1,1]^{[0,1]})^{k+1}$ and $\ell$ given by $y\mapsto\langle y,y',y'',\ldots y^{(k)}\rangle$ .
II Edit. The sufficient condition I gave above is of "extrinsic nature", and as such probably not in the spirit requested in the original question. An "intrinsic" condition, which is (probably) "simple", and in the line already suggested above in the first answer and in the comments, is that for any sequence $\langle x_i:i\in\mathbb N_0\rangle$ in $V$ satisfying $\lbrace 2^{i+2}(x_i-x_{i+1}):i\in\mathbb N_0\rbrace\subseteq B$ , there be $x\in V$ with $\lbrace 2^i(x_i-x):i\in\mathbb N_0\rbrace\subseteq B$ .
However, obviously this is not very practical to be verified in concrete situations. Basing on my experience and intuition, I would generally say that "extrinsic" conditions probably are more convenient than "intrinsic" ones. So, I think the question is good, but the restriction put there on the direction for searching for the answer is wrong. In practice, when one constructs (prospective new) Banach spaces, there is often some surrounding "larger" topological vector space where the new spaces will be continuously injected. In view of this, it is natural to look for extrinsic conditions.