Do inequalities that hold for infinite sums hold for integrals too?

As requested, there are some thoughts on the topic. If we have to pass from the properties of the sum to that of the Lebesgue integral, it's always worth to show that the property we are interested in does hold at least for the simple functions.

What does hold. Assume that $\{E_j\}_j$ is a countable measurable partition of $\Bbb R$ and consider a collection of non-negative functions $\{f_i\}_i$ $$ f_{i,j}(x) = \sum_{j} a_{ij}\cdot1_{E_j}(x) $$ so that all functions are simple and are given on the same partition. Note that $f:=T(\{f_i\}_i)$ is $$ f(x) = \sum_{j} T(\{a_{ij}\}_i)\cdot 1_{E_j}(x) $$ as you can check when substitute each single $x\in E_j$ into $T(\{f_i\}_i)$. As a result we have $$ \int f_i\mathrm d\lambda = \sum_j a_{ij} \lambda(E_j), \quad \int f\mathrm d\lambda = \sum_j T(\{a_{ij}\}_i) \lambda(E_j) $$ and if we put $x_{ij} = a_{ij}$ and $c_j = \lambda(E_j)$ then by the property of $T$ that is given to us: $$ T\left(\left\{\int f_i\mathrm d\lambda\right\}_i\right) = T\left(\left\{\sum_j a_{ij} \lambda(E_j)\right\}_i\right)\leq\sum_j \lambda(E_j) T(\{a_{ij}\}_i) = \int T(\{f_i\}_i)\mathrm d\lambda. $$

What holds in addition. If we are initially given simple functions on different (finite) partitions, then starting from $(\{E^i_j\}_j)_i$ - a countable collection indexed by $i\in \Bbb Z^+$ of finite partitions $E^i = \{E_{ij}\}_j$, there exists some countable partition $\{F_k\}_k$ which agrees with all of them. To show this, we first note that there is a countable partition $F^{1,2}$ underlying both $E^1$ and $E^2$, thus it holds for any finite collection $J\subset \Bbb Z^+$ of countable partitions - let us call the corresponding underlying partition by $F^J$. Now, if I am correct, $F$ contains at most as many elements as $\bigcup_n F^{1,\dots,n}$ and the latter is countable. Thus, our previous assumption is not restrictive and hence your claim does hold under assumption you have provided for all collections of simple functions.

What does hold under additional assumptions. Now, to pass from simple functions to measurable ones we can use the pointwise approximation by simple functions as in the construction of the Lebesgue integral. However, that requires interchanging $\sup_n$ and $T$ in the following way:

if $\{(y^n_i)_n\}_i$ is a countable collection of monotonically non-decreasing sequences $y_i = (y_i^0,y_i^1,\dots)$, then $$ \lim_n T(\{y_i^n\}_i) = T(\{\lim_n y^n_i\}_i). \tag{1} $$ Under such condition your inequality shall also hold for arbitrary measurable functions. I am not sure, whether $(1)$ is already implied by the assumptions raised in OP, so if you want to find a counterexample, I guess you shall look into cases when $(1)$ does not hold.

By the way, under the assumptions on the additivity of $T$ a condition similar to $(1)$ is completely characterized, and is known to hold exactly for transition kernels on $\Bbb R$, see [Proposition 1.3, Chapter 1] in Revuz, "Markov Chains".

That's pretty much what I could say, so hopefully it helps a bit.

A positive kernel on $\Bbb R$ is a function $K:\Bbb R\times \mathfrak B(\Bbb R)\to [0,\infty)$ such that $K(x,\cdot)$ is a positive measure for all $x$ and such that $K(\cdot,B)$ is a measurable function for all measurable $B$. Let $\mathfrak B_+(\Bbb R)$ denote the space of all non-negative measurable functions. Then $K$ defines an operator on that space by $$ Kf(x) = \int f(y)K(x,\mathrm dy). $$

Proposition: (Revuz) An additive and homogeneous operator $V$ on $\mathfrak B_+(\Bbb R)$ is associated with some positive kernel if and only if for every increasing sequence $\{f_n\}\subset \mathfrak B_+(\Bbb R)$ it holds that $$ V(\lim_n f_n) = \lim_n V(f_n). $$