Is there a nice characterization of those categories in which $a \subseteq b$ implies $a \leq b$? Given arrows $a:A \rightarrow Y$ and $b : B \rightarrow Y$, define that:


*

*$a \subseteq b$ iff for all $Z$ and all $g,g' : Y \rightarrow Z$, we have:


$$gb=g'b \rightarrow ga=g'a$$


*$a \leq b$ iff there exists $u : A \rightarrow B$ such that $a=bu.$


Then $a \leq b$ implies $a \subseteq b$, but not conversely. (By the way, does the relation $\subseteq$ have a conventional name, and is it discussed anywhere?)
Anyway, here's my question:

Question. Is there a nice characterization of those categories in which $a \subseteq b$ implies $a \leq b$ for all arrows $a$ and $b$ having a common codomain? I am especially looking for something that is useful for determining which categories of algebraic structures have this property. (Perhaps they all have it.)

 A: I think Condition 2 is the wrong condition to look at unless you put restrictions on $a$ and $b$ (e.g. that they are monomorphisms). Assume throughout the rest of this answer that the categories involved have finite limits and colimits. 
Condition 1 can be rephrased as follows. There is a universal object $Z$ equipped with two maps $g, g'$ such that $g \circ b = g' \circ b$: it is the pushout of the diagram $Y \xleftarrow{b} B \xrightarrow{b} Y$, also known as the cokernel pair $\text{coker}(b)$. It is the correct substitute for the cokernel in non-abelian settings: in particular, it is trivial iff $b$ is an epimorphism. The first condition can then be restated as the condition that $ga = g'a$ where $g, g'$ are these two universal maps, or equivalently that $a$ factors through the equalizer $\text{eq}(g, g')$. 
The equalizer of the cokernel pair is also called the regular image $\text{im}(b)$; it gives a factorization $B \to \text{im}(b) \to Y$ of $b$ where the map $\text{im}(b) \to Y$ is a regular monomorphism, and $\text{im}(b)$ is universal with this property. Hence we can restate the two conditions as follows:


*

*$a$ factors through $\text{im}(b) \to Y$. 

*$a$ factors through $B \to \text{im}(b) \to Y$.


Condition 1 is equivalent to the condition that the regular image $\text{im}(a) \to Y$ of $a$ factors through $\text{im}(b)$, which means it is really a condition on regular monomorphisms. In any category where regular monomorphisms are the same as monomorphisms, Condition 1 therefore reproduces the usual notion of containment of subobjects. (But regular monomorphisms already don't agree with monomorphisms for commutative rings.)
In the same way that there is a universal way to test Condition 1, there is a universal way to test the relationship between Condition 1 and Condition 2: taking $a$ to be the inclusion $\text{im}(b) \to Y$ itself shows that the two conditions are equivalent (for a fixed $b$) iff the map $B \to \text{im}(b)$ has a right inverse. Unfortunately there's no reason for this to be true in general even in an extremely nice category. 
Example. Let the ambient category be the category of abelian groups (about as nice as it gets!) and let $b : \mathbb{Z}_4 \to \mathbb{Z}_2$ be the obvious quotient. The regular image of $b$ is all of $\mathbb{Z}_2$, but the  quotient map $\mathbb{Z}_4 \to \mathbb{Z}_2$ does not have a right inverse; equivalently, the short exact sequence
$$0 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 0$$
does not split.
More generally, in an abelian category the map $B \to \text{im}(b)$ is an epimorphism and the regular image of any epimorphism is its codomain, so asking that every map of the form $B \to \text{im}(b)$ have a right inverse is equivalent to asking that every short exact sequence splits, which is equivalent to asking that the abelian category is semisimple. 
On the other hand, Conditions 1 and 2 are clearly equivalent if the map $B \to \text{im}(b)$ is an isomorphism, or equivalently if $b$ is a regular monomorphism. 
If you're interested in pursuing this further then I'd encourage you to learn more about your various options for defining images, since the intuition you seem to be trying to capture is "the image of $a$ is contained in the image of $b$." 
