Suppose $(\mathcal{E}, \mathcal{M})$ is a factorisation system for a category $\mathbf{C}$, $e : C \to C'$ is in $\mathcal{E}$ and $C''$ is an object in $\mathbf{C}$. Is it true that $$ e + \operatorname{id}_{C''} : C + C'' \to C' + C'' $$ is also in $\mathcal{E}$?
I somehow feel like this should be true, but cannot come up with a proof.
Yes, it is true, and in fact it is a special case of a more general fact : the class $\mathcal{E}$ is closed under colimits in the category of arrows. In other words, given two diagrams $F,F':\mathcal{K}\to \mathcal{C}$ and a natural transformation $\epsilon:F\Rightarrow F'$ where every component is in $\mathcal{E}$, the induced arrow $e:\operatorname{colim}F\to \operatorname{colim}F'$ is again in $\mathcal{E}$.
To prove this, it is enough to prove that $e$ is orthogonal to every arrow $m\in \mathcal{M}$. Given a commutative square $$\require{AMScd}\begin{CD}\operatorname{colim}F @>{e}>> \operatorname{colim}F'\\ @V{u}VV @VV{v}V\\ A@>>{m}> B, \end{CD}$$ precomposing with the canonical maps $\sigma_K:F(K)\to \operatorname{colim}F$ gives us for each object $K$ in $\mathcal{K}$ a commutative square $$\require{AMScd}\begin{CD}F(K) @>{\epsilon_K}>> F'(K)\\ @V{u\sigma_K}VV @VV{v\sigma'_K}V\\ A@>>{m}> B. \end{CD}$$ Then the orthogonality of $\epsilon_K$ and $m$ gives a unique map $\tau_K:F'(K)\to A$ such that $\tau_K\epsilon_K=u\sigma_K$ and $m\tau_K=v\sigma_K'$.
Now for any $f:K\to K'$ in $\mathcal{K}$ we have $$\tau_{K'}F'(f)\epsilon_K=\tau_{K'}\epsilon_{K'}F(f)=u\sigma_{K'}F(f)=u\sigma_K$$ and $$m\tau_{K'}F'(f)=v\sigma_{K'}'F'(f)=v\sigma'_K,$$ thus the uniqueness implies that $\tau_{K'}F'(f)=\tau_K$. Then by the universal property of $\operatorname{colim}F'$ there exists a unique $t:\operatorname{colim}F'\to A$ such that $t\sigma'_K=\tau_K$ for all $K$. To prove the identities $te=u$ and $mt=v$, it is enough to observe that $$te\sigma_K=t\sigma_{K}'\epsilon_K=\tau_K\epsilon_K=u\sigma_K$$ and $$mt\sigma'_K=m\tau_K=v\sigma'_K.$$ Conversely, these equalities show that for any map $t'$, $t'e=u$ and $mt'=v$ imply that $t'\sigma'_K=\tau_K$, and thus that $t'=t$, which proves the uniqueness.